ejercicios fisica cinematica

8/17/2019 EJERCICIOS FISICA CINEMATICA

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Profesor de Física WALTER PEREZ TERREL TEMA: Cinemática

1

CALCULO DE LA DERIVADA.OPERADOR DERIVADA.

A. Determinar la primera derivada de las siguientes funciones. Determinarla velocidadpara: t=2 segundos.

1. X (t)= -5t V (t)= -5Para t=2: V (2)= -5

2. X (t)= t V (t)= Para t=2: V (2)=

3. X (t)= ! V (t)= "Para t=2: V (2)= "

4. x ( t )=

−t 3

v ( t )=−13

Parat =2 v ( t )=−13

5. ( t )=

t

2v ( t )=1

2

Para ( t )=2 v ( t )=1

2

6. X (t)= -!t# V (t)= -$2t2

Para t=2: V (2)= -$2(!) = -!%

7. X (t)=−t 3

3 V (t)= -t2

Para t=2: V (2)= -!

8. X (t)= #t2 V (t)= tPara t=2: V (2)= (2) = $2

9. X (t)=t 2

2 V (t)= t

Para t=2: V (2)= 2

10. X (t)= t ! V (t)= t#

!


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Para t=2: V (2)= %

11. X (t)= !t! V (t)= $t#

Para t=2: V (2)= $(%) = $2%

12. X (t)=t 4

4 V (t)= t#

Para t=2: V (2)= %

13. X (t)= -2t5 V (t)= -$"t!

Para t=2: V (2)= -$"($) = -$"

14. X (t)= t! & #t2 & 5t V (t)= !t# & t - 5Para t=2: V (2)= !(%)-(2)-5 =$5

15. X (t)= 2t!'#t2-5t'% V (t)= %t#'t-5 Para t=2: V(2)= %(%)'(2)-5=$

16. X (t)= 2t#-5t2-5t-2 V (t)= t2-$"t-5 Para t=2: V(2)= (!)-$"(2)-5=-$

17. X (t)= -5t

-$

V (t)= 5t

-2

Para t=2: v (2 )=5

4

18. X (t)= -#t-2 V (t)= t-#

Para t=2 v (2)=6

8

19. X (t)= -2t-# V (t)= t-!

Para t=2: V (2)=6

16

20. X (t)= √ t V (t)=1

2√ t

Para t=2: V (2)=1

2√ 2

21. X (t)= √ t +3 V (t)=

1

2√ t +3


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Para t=2: V (2)=1

2√ 5

22. X (t) =√ t 2+3

V (t) =

2 t

2√ t 2

+3

Para t=2: V (2)=2

√ 7

23. X (t)= √ t 2+3 t V (t)=

2t +3

2√ t 2+3

Para t=2: V (2)=7

2√ 7

24. X (t) = √ t 2+3 t +5 V (t) =

2 t +3

2√ t 2+3 t +5

Para t=2: V (2) =7

2√ 15

25. X (t) = sen (t) V (t) = cos (t)Para t=2: V (2) = cos (2)

26. X (t) = -5sen (t) V (t) = -5cos (t)Para t=2: V (2) = -5cos (2)

27. X (t) =-! sen (2t) V (t) = -!cos (2t) 2Para t=2: V (2)=-% cos (!)

28. X (t) = -%sen (#t) V (t) = -%cos (#t) #Para t=2: V (2)= -2!cos ()

29. X (t) = -%sen (#t2) V (t) = -%cos (#t2)t

Para t=2: V (2)= -2cos ($2)30. X (t)= %sen (#t'2) V (t)= %cos (#t'2)#

Para t=2: V (2)=2!cos (%)

31. X (t) = %sen (#t2'2) V (t) = %cos (#t2'2)tPara t=2: V (2)= 2cos (%)

32. X (t) =2 sen (#t2'2t)V (t) =2 cos (#t2'2t) t'2Para t=2: V (2)= 2%cos ($!)

33. X (t) =-! sen (#t2-2t'5)V (t) =-! cos (#t2-2t'5)t'2


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Para t=2: V (2)= -5cos ($)

34. X (t) = sen ( √ t ) V (t) = cos ( √ t )

Para t=2: V (2)= cos ( √ 2 )

35. X (t) = sen ( √ t 2+2 t +1 )

V (t)=cos ( √ t 2+2 t +1 ) 2t'2

2 √ t 2+2 t +1

Para t=2: V (2)= cos (#)

36. X (t) = c os√ t 2+2 t +1 )

V (t) =-sen ( √ t 2+2 t +1 ) 2t'2

2 √ t 2+2 t +1

Para t=2: V (2)= -sen (#)

37. X (t) = cos ( √ t ) V (t) = -sen ( √ t )1

2√ t

Para t=2: V (2) = -sen ( √ 2 )

2√ 2

38. X (t) = cos (t) V (t) = -sen (t)Para t=2: V (2) = -sen (2)

39. X (t) = cos (2t) V (t) = -sen (2t) 2Para t=2: V (2) = 2-sen (!)

40. X (t) = cos (#t2) V (t) = -sen (#t2)t

Para t=2: V (2) = -$2sen ($2)

41. X (t) = cos ( √ t ) V (t) = -sen ( √ t )1

2√ t

Para t=2: V (2) =-sen ( √ 2 )1

2√ 2

42. X (t) = cos (#t2'2t)V (t) = -sen (#t2'2t) t'2

Para t=2: V (2) = -$!sen ($)

43. X (t) = t!'#t2 V (t) = !t#'t


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Para t=2: V (2) = !(%) '(2) =!!

44. X (t)= t5 V (t)=5 t!

Para t=2: V (2)=5($)= %"

45. X (t) = t't! V (t) = t5'!t#Para t=2: V (2)= (#2)'!(%)=22!

46.23

)(46

t t t X −=

X ' ( t )=1

3.6 t

5−1

2.4 t

3

X ' ( t )=2 t 5−2t 3

V ( t )= X ' ( t )=2 t 5−2t 3

V (2 )=2(2)5−2(2)3

V (2 )=64−16

V (2 )=48m .s−1

47.46

)(

46

t t t X +=

X ' ( t )=1

6.6 t

5−1

4.4 t

3

X ' (t )=t 5−t 3

V ( t )= X ' ( t )=t 5−t 3

V (2 )=(2)5

−(2)

3

V (2 )=32−8

V (2 )=24m .s−1

48.

)(.)( t Sent t X −=

X ' (t )= (−t )' sen ( t )+(−t )cos (t )

X ' (t )=−sen (t )−tcos (t )


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V ( t )= X ' ( t )=−sen (t )−tcos ( t )

V (2 )=−sen (2 )−2cos (2 )

V (2 )=−0.034−2(0.999)

V (2 )=2.032m.s−1

49.

)(.)( 2 t Sent t X −=

X ' (t )= (−t 2 )' sen (t )+ (−t 2 )cos (t )

X ' (t )=−2 tsen ( t )−t 2 cos (t )

V ( t )= X ' ( t )=−2 tsen ( t )−t 2cos ( t )

V (2 )=−4 sen (2 )−4cos (2)

V (2 )=−4 (0.034 )−4(0.999)

V (2 )=−4.132m. s−1

50.

)2(.)( 2 t Sent t X −=

X ' (t )= (−t 2 )' sen (2 t )+(−t 2 )cos (2t )(2)

X ' (t )=−2 tsen (2 t )−2 t 2 cos (2 t )

V ( t )= X ' ( t )=−2 tsen (2 t )−2t 2cos (2t )

V (2 )=−4 sen (4 )−8cos (4 )

V (2 )=−4 (0.069 )−8(0.997)

V (2 )=8.252m.s−1

51.

)(.)( t Cost t X =

X ' ( t )= ( t )' cos (t )+( t ) (−sen (t ))

X ' (t )=cos (t )−tsen ( t )

V ( t )= X ' ( t )=cos (t )−tsen (t )

V (2 )=cos (2 )−2sen (2 )

V (2 )=0.999−2(0.034)


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V (2 )=0.931m.s−1

52.)2(.3)( t Cost t X =

X ' (t )= (3 t )' cos (2t )+ (3 t ) (−sen (2 t ))2

X ' (t )=3cos (2t )−6 tsen (2t )

V ( t )= X ' ( t )=3cos (2 t )−6 tsen (2 t )

V (2 )=3cos (4 )−12 sen (4 )

V (2 )=3 (0.997)−12 (0.069)

V (2 )=2.163m.s−1

53.

)3(.)( t Cost t X =

X ' (t )= ( t )' cos (3 t )+( t ) (−sen(3t ))3

X ' (t )=cos (3 t )−3tsen (3 t )

V ( t )= X ' ( t )=cos (3 t )−3 tsen (3t )

V (2 )=3cos (6 )−6 sen (6 )

V (2 )=3 (0.994 )−6(0.104)

V (2 )=2.358m.s−1

54.

)(.)( 2 t Cost t X =

X ' (t )= (t 2 )' cos ( t )+( t 2 ) (−sen ( t ) )

X ' (t )=2tcos ( t )−t 2sen ( t )

V ( t )= X ' ( t )=2 tcos ( t )−t 2 sen ( t )

V (2 )=4 cos(2 )−4 sen (2 )

V (2 )=4 (0.999 )−4(0.034 )

V (2 )=3.86m. s−1


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*. Determinar la segunda derivada de las siguientes funciones. Determinar laA+,,A+/01 para: t = # segundos.

$.

2.5)( t t X −=

X ' ( t )=−10t

X ' ' (t )=−10

a (3 )= X ' ' (t )=−10

a (3 )=−10m. s−2

2.t t X .4)( =

X ' ( t )=4

X ' ' ( t )=0

a (3 )= X ' ' ( t )=0

a (3 )=0m.s−2

#.

3.4)( t t X −=

X ' (t )=−12 t 2

X ' ' ( t )=−24 t

a (t )= X ' ' ( t )=−24 t

a (3 )=−24 (3)

a (3 )=−72m.s−2

!.

4.3)( t t X =

X ' (t )=12 t 3

X ' ' ( t )=36 t 2

a (t )= X ' ' (t )=36 t 2

a(3 )=36

(3

)

2

a (3 )=324 m. s−2


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5.

5.4)( t t X =

X ' (t )=20t 4

X ' ' ( t )=80 t 3

a (t )= X ' ' ( t )=80 t 3

a (3 )=80(3)3

a (3 )=2160m .s−2

. t t t t X .5.3)( 23

−−= X

' (t )=3t 2−6 t −5

X ' ' ( t )=6 t −6

a (t )= X ' ' ( t )=6 t −6

a (3 )=6(3)−6

a (3 )=12m . s−2

.8.5.3.2)( 24 +−+= t t t t X

X ' (t )=8 t 3+6 t −5

X ' ' ( t )=24 t 2+6

a (t )= X ' ' ( t )=24 t 2+6

a (3 )=24(3)2+6

a (3 )=222m .s−2

%.2.5.5.2)( 23 −−−= t t t t X

X ' (t )=6 t 2−10 t −5

X ' ' ( t )=12 t −10

a (t )= X ' ' ( t )=12t −10


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a (3 )=12(3)−10

a (3 )=16m.s−2

.t

t X 5)( −=

X ' (t )=5 t −2

X ' ' ( t )=−10 t −3

a (t )= X ' ' ( t )=−10 t −3

a (3 )=−10(3)−3

a (3 )=−0.37m .s−2

$".

2

3)(

t t X −=

X ' (t )=6 t −3

X ' ' ( t )=−18 t −4

a (t )= X ' ' ( t )=−18 t −4

a (3 )=−18(3)−4

a (3 )=−0.22m. s−2

$$.

3

2)(

t t X −=

X ' (t )=6 t −4

X ' ' ( t )=−24 t −5

a (t )= X ' ' ( t )=−24 t −5

a (3 )=−24 (3)−5

a (3 )=−0.098m .s−2

$2.

)()( t Sent X =

X ' (t )=cos (t )


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X ' ' ( t )=−sen (t )

a (t )= X ' ' ( t )=−sen(t )

a (3 )=−sen(3)

a (3 )=−0.052m. s−2

$#.)(.5)( t Sent X −=

X ' (t )=−5cos (t )

X ' ' ( t )=5 sen (t )

a (t )= X ' ' ( t )=5sen( t )

a (3 )=5 sen(3)

a (3 )=0.261m .s−2

$!.).2(.4)( t Sent X −=

X ' (t )=−8cos (2 t )

X ' ' ( t )=16 sen (2t )

a (t )= X ' ' ( t )=16sen(2t )

a (3 )=16 sen(6)

a (3 )=0.261m .s−2

$5.).3(.8)( t Sent X −=

X ' (t )=−24cos (3t )

X ' ' ( t )=72 sen (3 t )

a (t )= X ' ' ( t )=72sen(3 t )

a (3 )=72 sen(9)

a (3 )=11.263m.s−2


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$.)2.3(.8)( += t Sent X

X ' (t )=24cos (3 t +2)

X ' ' ( t )=−72 sen(3 t +2)

a (t )= X '' (t )=−72sen(3 t +2)

a (3 )=−72sen(11)

a (3 )=13.738m .s−2

$.)()( t Sent X =

X ' (t )=

cos (√ t )

2√ t

X ' ' ( t )=

−√ t sen (√ t )−cos (√ t )4 t √ t

a (t )= X ' ' ( t )=

−√ tsen (√ t )−cos (√ t )4 t √ t

a (3 )=−1.73sen (√ 3)−cos (√ 3)12√ 3

a (3 )=−0.052−0.99920.76

a (3 )=−0.051m. s−2

$%.)()( t Cost X =

X '

(t )=−sen(t )

X ' ' ( t )=−cos ( t )

a (t )= X ' ' ( t )=−cos ( t )

a (3 )=−cos (3)

a (3 )=−0.998m .s−2

$.).2()( t Cost X =


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X ' (t )=−2 sen (2 t )

X ' ' ( t )=−4 cos (2t )

a (t )= X ' ' ( t )=−4cos (2 t )

a (3 )=−4cos (4 )

a (3 )=−3.002m.s−2

2".).3()( 2t Cost X =

X ' (t )=−6 t sen(3t 2)

X ' '

( t )=−6 sen (3t 2

)−36 t 2

cos (3 t 2

)

a (t )= X ' ' ( t )=−6 sen (3 t 2)−36 t 2cos (3 t 2)

a (3 )=−6 sen (27)−324cos (27)

a (3 )=−6 (0.453)−324(0.891)

a (3 )=−291.404m . s−2

2$.)()( t Cost X =

X ' (t )=

−sen (√ t )

2√ t

X ' ' ( t )=

−√ t cos (√ t )+sen (√ t )4 t √ t

a (t )= X ' ' ( t )=−√ t cos (√ t )+sen (√ t )

4 t √ t

a (3 )=−1.73cos (√ 3 )+sen (√ 3)

12√ 3

a (3 )=−1.729−0.030

20.76

a (3 )=−0.085m .s−2

22.

t t X =)(


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X ' (t )=

1

2√ t

X ' ' ( t )= −1

4 t √ t

a (t )= X ' ' ( t )= −14 t √ t

a (3 )= −1

12√ 3

a (3 )= −120.76

a (3 )=−0.048m .s−2

2#.3)( += t t X

X ' (t )= 1

2√ t +3

X ' ' ( t )= −1

4 (t +3)√ t +3

a (t )= X ' ' ( t )= −14(t +3)√ t +3

a (3 )= −1

24√ 6

a (3 )= −158.79

a (3 )=−0.017m .s−2

C. Determinar la tercera derivada de las siguientes funciones. Determinar la+,,/DAD para: t = $ segundo.

$.2

)(2t

t X =

X ' (t )= t

X ' ' ( t )=1

X ' '' ( t )=0


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C ( t )= X ' ' ' ( t )=0

C (1)=0m .s−2

2.3

)(3

t t X =

X ' (t )= t 2

X ' ' ( t )=2 t

X ' '' ( t )=2

C ( t )= X ' ' ' (t )=2

C (1)=2m .s−2

#.8

)(4t

t X =

X ' (t )= t

3

2

X ' ' ( t )=3 t

2

2

X ' '' ( t )=3 t

C ( t )= X ' ' ' (t )=3 t

C (1)=3m .s−2

!.).3()( t Sent X =

X

' (t

)=

3cos (3

t ) X

' ' ( t )=−9 sen(3 t )

X ' '' ( t )=−27 cos (3 t )

C ( t )= X ' ' ' (t )=−27cos(3 t )

C (1)=−27cos (3)

C (1)=−26.962m. s−2


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5.

).4()( t Sent X =

X ' (t )=4cos (4 t )

X ' ' ( t )=−16 sen (4 t )

X ' '' ( t )=−64cos (4 t )

C ( t )= X ' ' ' (t )=−64cos(4 t )

C (1)=−64cos (4 )

C (1)=−63.844m .s−2

.)()( π

+= t Sent X

X ' (t )=cos (t +π )

X ' ' ( t )=−sen (t +π )

X ' ' ' ( t )=−cos (t +π )

C ( t )= X ' ' ' ( t )=−cos(t +π )

C (1)=−cos(1+π )

C (1)=0.999m .s−2

.)2()( π += t Sent X

X ' (t )=2cos (2 t +π )

X ' ' ( t )=−4 sen(2t +π )

X ' '' ( t )=−8cos (2t +π )

C ( t )= X ' ' ' ( t )=−8cos(2 t +π )

C (1)=−8cos (2+π )

C (1)=−7.995m.s−2

%.)()( 2 π += t Sent X

X ' (t )= tcos(t 2+π )

X ' ' ( t )=−t 2 sen(t 2+π )


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X ' '' ( t )=−t 3 cos (t 2+π )

C ( t )= X ' ' ' (t )=−t 3 cos(t 2+π )

C (1)=−

13cos

(1

+π )

C (1)=0.999m .s−2

.

).2()( π π += t Sent X

X ' (t )=2πcos (2πt +π )

X ' ' ( t )=−4 π 2 sen(2πt +π )

X

' ''

( t )=−8π 3

cos(2πt +π )

C ( t )= X ' ' ' (t )=−8π 3cos (2πt +π )

C (1)=−8 (3.14 )3(−1)

C (1)=247.7m .s−2

$".).2()( t Cost X =

X ' (t )=−2 sen (2 t )

X ' ' ( t )=−4 cos(2 t )

X ' '' ( t )=8 sen (2 t )

C ( t )= X ' ' ' (t )=8 sen(2 t )

C (1)=8 sen(2)

C (1)=0.279

m .s

−2

$$.).3()( t Cost X =

X ' (t )=−3 sen(3t )

X ' ' ( t )=−9cos (3t )

X ' '' ( t )=27sen (3 t )

C ( t )= X ' ' ' ( t )=27 sen(3 t )


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C (1)=27 sen(3)

C (1)=1.413m .s−2

$2.).4()( t Cost X =

X ' (t )=−4 sen (4 t )

X ' ' ( t )=−16cos(4 t )

X ' '' ( t )=64 sen(4 t )

C ( t )= X ' ' ' (t )=64 sen(4 t )

C (1)=64 sen(4)

C (1)=4.464m. s−2

$#.

)()( π += t Cost X

X ' (t )=−sen (t +π )

X ' ' ( t )=−cos( t +π )

X ' '' ( t )=sen ( t +π )

C ( t )= X ' ' ' ( t )=sen(t +π )

C (1)=sen(1+π )

C (1)=0.017 m. s−2

$!.( )t Cost X .2)( π =

X ' (t )=−2 πsen(2 πt )

X ' ' ( t )=−4 π 2 cos(2πt )

X ' '' ( t )=8π 3 sen (2πt )

C ( t )= X ' ' ( t )=8π 3 sen(2 πt )

C (1)=8 (3.14)3(0)

C (1)=0m .s−2


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$5.( )π π −= t Cost X .2)(

X ' (t )=−2 πsen(2 πt −π )

X ' ' ( t )=−4 π 2 cos(2πt −π )

X ' '' ( t )=8π 3 sen (2πt −π )

C ( t )= X ' ' ( t )=8π 3 sen(2 πt −π )

C (1)=8 (3.14)3(0)

C (1)=0m .s−2

$.

+= 2.2)( π

π

t Cost X

X ' (t )=−2 πsen(2πt + π 2 )

X ' ' ( t )=−4 π 2 cos(2πt + π 2 )

X ' '' (t )=8π 3 sen(2 πt + π 2 )

C ( t )= X ' ' ( t )=8π 3 sen(2 πt + π 2 )C (1)=8 π 3 sen(2π −π 2 )C (1)=8 (3.14)3(1)

C (1)=247.7m .s−2

$.

−=

2.2)(

π π t Cost X

X ' (t )=−2 πsen(2πt −π 2 )

X ' ' ( t )=−4 π 2 cos(2πt −π 2 )

X

' ''

( t )=8π 3

sen

(2 πt −

π

2

)


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C ( t )= X ' ' ( t )=8π 3 sen(2 πt − π 2 )C (1)=8 (3.14)3(−1)

C (1)=−247.7m. s−2

18.

).4()( t Cost X =

v ( t )=dx

dt =−4sin4. t

a ( t )=dv

dt =−16cos 4 t

celeridad=da

dt =64sin4 t =64sin 4=4.46

19.

)()( π += t Cost X

t +π (¿)

v ( t )=dx

dt =−sin ¿

a ( t )=

dv

dt =−cos ( t +π )

celeridad=da

dt =sin (t +π )

20.

( )t Cost X .2)( π =

v ( t )=dx

dt =−2π sin (2 π . t )

a ( t )=dvdt =−4 π 2cos 2π .t

celeridad=da

dt =8π 3 sin 2π . t

21.

( )π π −= t Cost X .2)(

v ( t )=dx

dt =−2π sin (2πt −π )


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a ( t )=dv

dt =−4 π 2cos (2πt −π )

celeridad=da

dt =8π 3 sin(2πt −π )

22.

+=

2.2)(

π π t Cost X

v ( t )=dx

dt =−2π sin (2πt +

π

2)

a ( t )=dvdt =−4 π 2cos(2 πt + π 2 )

celeridad=da

dt =8π 3(2πt + π 2 )

−=

2.2)(

π π t Cost X

2#.

v ( t )=dxdt =−¿−2π sin(2πt −π 2 )

a ( t )= dvdt =−4 π 2cos(2 πt − π 2 )

celeridad=da

dt =8π 3(2πt − π 2 )

EJERCICIOS Y PROBLEMAS PROPUESTOS

ema: 34V/3/,14 1/D/3,16/41A(1ivel *7sico)

1. n cuerpo tiene la siguiente le8 del movimiento: X

(t)

23 4.t t

9 donde t semide en segundos 8 X se mide en metros. Determine la distancia ue recorreentre los instantes t = 2 s 8 t = 5 s.

Parat =2 s : x (t )=3+4 (2 )+42=27m

Parat =5 s : x (t )=3+4 (5 )+52=48m

∴distancia=⌈ ∆ t ⌉=48−27=21m


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22

2. a posici;n de un partne mediante la le8:2 3( ) 4 5. 6. 2. X t t t t = + + +

9 donde t se mide en segundos 8 X en metros.Determine la velocidad de la part


23/84


23

B) Determine la velocidad en el instante t = s.c) C,n u instante la velocidad es nulaE

v ( t )=dx

dt =3 t 2−18

a)3¿¿

Parat =3 s :3¿

B) Parat =9 s :3 (9 )2−18=225m /s

c) v ( t )=0=3t 2−18

t =√ 6 s

6. 6e conoce la le8 del movimiento de una part


24/84


24



25/84


25

segundos 8 ?@ en metros.a) Determine la aceleraci;n en el instante t = $ s.B) Determine la aceleraci;n en el instante t = 5 s.c) C,n u instante la aceleraci;n es nulaE

v ( t )=dx

dt =6 t 2−24 t

a ( t )=dv

dt =12t

a) Parat =1 s :a (t )=12(1)=12m / s2

B) Parat =5 s :a ( t )=12 (5 )=60m/s2

c) a ( t )=0=12t

t =0



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26

a) Parat =2 s :a (t )=12 (2 )2−36=12m /s2

B) Parat =4 s :a (t )=12 (4 )2−36=156 m/ s2

c) a ( t )=0=12t 2−36

t =√ 3 s



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27

B) Determine la velocidad en el instante t = ! s.c) Determine la aceleraci;n media entre los instantes t = $ s 8 t = # s.

v ( t )=dx

dt =4 t 3−36 t

a) Parat =2 s :v (t )=4 (8 )−36 (2 )=−40m /s

B) Parat =4 s :v ( t )=4 (64 )−36 (4 )=112m /s

c) ⃗am=v f −v i

∆ t

Parat =1 s :v (t )=4 (1 )−36=−32m /s

Parat =3 s : v (t )=4 (27 )−36 (3 )=0

⃗am=322=16m / s2



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28

v ( t )=dx

dt =10 π .

t

4cos(π t

2

8+

π

2 )v ( t )=0=5π . t

2cos( πt

2

8+

π

2 )

19.

st 2= st 5= st 1=+onociendo la le8 del movimiento:

10.1223

)(23

+−−= t t t

t X

9 donde ?X se mide en metros 8 t se mide ensegundos. Determinar:a) Determinar la velocidad en cualuier instante de tiempo.B) C,n u instante la velocidad es nulaEc) C,n u posici;n su velocidad es nulaE

d) C+u7l es su recorrido 8 desplaFamiento entre st 0= 8

st 5= Ee) C+u7l es su velocidad media entre 8 GGGGGE

f) C+u7l es su aceleraci;n media entre 8 st 5=

Eg) Determinar la aceleraci;n en cualuier instante de tiempo.

H) C+u7l es su aceleraci;n en st 4=

Ei) C,n u instante la aceleraci;n es nulaE ) C,n u posici;n su aceleraci;n es nulaE

) v (t )= dx

dt = t 2−t −12

!) v (t )=0=t 2− t −12

t =4 s∧ t =−3 s

t no puede ser igual a−3 porue¿

el tiempono puede sernegativo ¿ ∴t =4 s

c)

10.1223

)(23

+−−= t t t t X

v ( t )=0cuando t =4 s

x ( t )=43

3−

42

2−12∗4+10=−24.6

") despla!amiento=∆ x

Parat =0: x ( t )=10


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29

Parat =5 s : x (t )=125

3−

25

2−60+10=−20.8

∴despla!amiento=∆ x=−20.8−10=−30.8m

#) ⃗vm=∆ x∆ t

Parat =1 s : x (t )=13−

1

2−12+10=−2.1

Parat =5 s : x (t )=−20.8

∴⃗vm=−20.8+2.1

4=−4.6m. s−1

$) ⃗am=

∆ v

∆ t

Parat =2 s :v (t )=4−2−12=−10m.s−1

Parat =5 s : v (t )=25−5−12=8m. s−1

∴⃗am=18

3

=6m .s−2

%) a=

dv

dt =2 t −1

&) t =4

a=2 (4 )−1=7m .s−2

') a=0=2t −1

t =1

2s

()

10.1223

)(23

+−−= t t t

t X

a=0cuandot =1

2s


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30

x (t )= 1

24−

1

8−6+10=3.9m

20. +onociendo la le8 del movimiento:

10.10

2

.3

3

)(23

+−−= t t t

t X

9 donde ?X se mide enmetros 8 t se mide en segundos. Determinar:a) Determinar la velocidad en cualuier instante de tiempo.B) C,n u instante la velocidad es nulaEc) C,n u posici;n su velocidad es nulaE

d) C+u7l es su recorrido 8 desplaFamiento entre st 0=

8 st 5=

e) C+u7l es su velocidad media entre st 1=

8 st 5=

E

f) C+u7l es su aceleraci;n media entre

st 2= 8

st 5=E

g) Determinar la aceleraci;n en cualuier instante de tiempo.

H) C+u7l es su aceleraci;n en st 4=


) v (t )=

dx

dt = t 2−3 t −10

!) v (t )=0=t 2−3t −10

t =5s∧ t =−2 s


el tiempono puede ser negativo¿ ∴t =5 s

c)

10.102

.3

3)(

23

+−−= t t t

t X

v ( t )=0cuandot =5 s

x ( t )=53

3−

3∗52

2−10∗5+10=−35.8m


Parat =0: x ( t )=10

Parat =5 s : x (t )=125

3−

3∗252

−50+10

¿−35.8m


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∴despla!amiento=∆ x=−35.8−10

¿−45.8m

#) ⃗vm=

∆ x

∆ t

Parat =1 s : x (t )=13−

3

2−10+10=−1.1

Parat =5 s : x (t )=−20.8

∴⃗vm=−20.8+1.1

4=−4.9m.s−1

$) ⃗am=

∆ v

∆ t

Parat =2 s :v (t )=4−3 (2)−10=−12m.s−1

Parat =5 s : v (t )=25−3(5)−10=0m.s−1

∴⃗am=12

3=4 m.s−2

%) a=

dv

dt =2 t −3

&) t =4 s

a=2 (4 )−3=5m .s−2

') a=0=2t −3

t =1.5s

()

10.102

.3

3)(

23

+−−= t t t

t X

a=0cuandot =1.5s

x ( t )=2724

−27

8−15+10=−7.25


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32


10.623

)(23

+−−= t t t

t X

9 donde ?X se mide enmetros 8 t se mide en segundos. Determinar:a) Determinar la velocidad en cualuier instante de tiempo.

B) C,n u instante la velocidad es nulaEc) C,n u posici;n su velocidad es nulaE


8 st 5=

E


8 st 5=

E

f) C+u7l es su aceleraci;n media entre st 2=

8 st 5=


H) C+u7l es su aceleraci;n en

st 4=


)

10.623

)(23

+−−= t t t

t X

v ( t )=dx

dt =t 2−t −6

!) v (t )

=0

=t 2

−t −

6

t =3s∧ t =−2 s



c)

10.623

)(23

+−−= t t t

t X

v ( t )=0cuandot =3s

x ( t )=33

3−

32

2−6 (3 )+10=−3.5m


Parat =0: x ( t )=10

Parat =5 s : x (t )=125

3

−25

2

−30+10=9.1m


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∴despla!amiento=∆ x=9.1−10=−0.9m

#) ⃗vm=

∆ x

∆ t

Parat =1 s : x (t )=1

3−1

2−6+10=3.8

Parat =5 s : x ( t )=9.1

∴⃗vm=9.1+3.8

4=3.2m. s−1

$) ⃗am=

∆ v

∆ t

Parat =2 s :v (t )=4−2−6=−4m. s−1

Parat =5 s : v (t )=25−5−6=14m .s−1

∴⃗am=18

3=6m .s−2

%)

a=dv

dt

=2 t −1

&) t =4

a=2 ( 4 )−1=7m .s−2

') a=0=2t −1

t =1

2s

()

10.623

)(23

+−−= t t t

t X

a=0cuandot =1

2s

x ( t )= 124

−1

8−6+10=3.9m


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34


10.182

.3

3)(

23

+−−= t t t

t X




8 st 5=

E


8 st 5=

E


8 st 5=

Eg) Determinar la aceleraci;n en cualuierinstante de tiempo.

H) C+u7l es su aceleraci;n en st 4= E

i) C,n u instante la aceleraci;n es nulaE ) C,n u posici;n su aceleraci;n es nulaE

)

10.182

.3

3)(

23

+−−= t t t

t X

v ( t )=dx

dt =t 2−3 t −18

!) v (t )=0=t 2−3t −18

t =6s∧t =−3 s



c)

10.182

.3

3)(

23

+−−= t t t

t X

v ( t )=0cuando t =6 s

x ( t )=63

3−

3∗62

2−18 (6 )+10=−80m


Parat =0: x ( t )=10

Parat =5 s : x (t )=1253 − 3∗252 −90+10=−75.8m


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35

∴despla!amiento=∆ x=−75.8−10=−85.8m

#) ⃗vm=

∆ x

∆ t

Parat =1 s : x (t )=13−3

2−18+10=−9.16

Parat =5 s : x ( t )=−75.8

∴⃗vm=−75.8+9.16

4=−16.6m. s−1

$)

⃗am=∆ v

∆ t

Parat =2 s :v (t )=4−6−18=−20m .s−1

Parat =5 s : v (t )=25−15−18=−8m.s−1

∴⃗am=12

3=4 m.s−2

%) a=dv

dt =2 t −3

&) t =4

a=2 (4 )−3=5m .s−2

') a=0=2t −3

t =3

2s

()

10.182

.3

3)(

23

+−−= t t t

t X

a=0cuandot =3

2s

x ( t )=2724

−27

8−27+10=−19.25m


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36


10.16.33

)( 23

+−−= t t t

t X




8 st 10=

E


8 st 10=

E


8 st 10=



st 4=


)

10.16.33

)( 23

+−−= t t t

t X

v ( t )=dx

dt =t 2−6 t −16

!) v ( t )=0=t 2−6 t −16

t =8 s∧t =−2 s


el tiempono puede ser negativo¿

∴t =6 s

c)

10.16.3

3

)( 23

+−−= t t t

t X


x ( t )=63

3−

3∗62

2−16 (8 )+10=−100m


Parat =5 : x ( t )=

125

3 −3 (25)−16 (5 )+10=−103.3m


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37

Parat =10 s : x ( t )=1000

3−3 (100)−160+10=−116.6m

∴despla!amiento=∆ x=−116.6+103.3=−13.3m

#) ⃗vm=

∆ x

∆ t

Parat =6 s : x (t )=216

3−3 (36 )−16 (6 )+10=−122m

Parat =10 s : x ( t )=−116.6m

∴⃗vm=

−116.6+1224 =1.35m. s

−1

$) ⃗ am=

∆ v

∆ t

Parat =6 s : v ( t )=36−36−16=−16m .s−1

Parat =10 s :v ( t )=100−60−16=24m .s−1

∴⃗am=40

3=13.3m .s−2

%) a=

dv

dt =2 t −6

&) t =4

a=2 ( 4 )−6=2m .s−2

') a=0=2t −6

t =3s

()

10.16.33

)( 23

+−−= t t t

t X

a=0cuandot =3s


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38

x ( t )=9−27−48+10=−56m

24. +onociendo la le8 del movimiento:5)( 23 +−−= t t t t X

9 donde ?X se mide en metros8 t se mide en segundos. Determinar:

a) Determinar la velocidad en cualuier instante de tiempo.B) C,n u instante la velocidad es nulaEc) C,n u posici;n su velocidad es nulaE


8 st 2=

E


8 st 5=

E


8 st 5=


H) C+u7l es su aceleraci;n en st 5= Ei) C,n u instante la aceleraci;n es nulaE ) C,n u posici;n su aceleraci;n es nulaE

)

5)( 23 +−−= t t t t X

v ( t )=dx

dt =3 t 2−2 t −1

!) v ( t )=0=3t 2−2 t −1

t =1s∧t =−13

s

t no puede ser igual a−1

3 porue

¿

el tiempono puede sernegativo¿ ∴t =1 s

c)5)(

23

+−−= t t t t X


x ( t )=1−1−1+5=4m


Parat =0: x ( t )=5

Parat =2 s : x (t )=8−4−2+5=7m


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39

∴despla!amiento=∆ x=7−5=2m

#) ⃗vm=

∆ x

∆ t

Parat =3 s : x (t )=27−9−3+5=20

Parat =5 s : x (t )=125−25−5+5=100

∴⃗vm=100−20

2=40m . s−1

$) ⃗am=

∆ v

∆ t

Parat =2 s :v (t )=3 (4 )−2 (2 )−1=7 m.s−1

Parat =5 s : v (t )=3(25)−2(5)−1=64 m.s−1

∴⃗am=57

3=19m .s−2

%)

a=dv

dt

=6t −2

&) t =5 s

a=6 (5 )−2=28m.s−2

') a=0=6 t −2

t =1

3s

()

5)( 23 +−−= t t t t X

a=0cuandot =1

3s

x ( t )= 127

−1

9−

1

3+5=4.5m


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40


5.62

.5

3)(

23

++−= t t t

t X




8 st 4=

E


8 st 5=

E


8 st 5=



st 5,2=


)

5.62

.5

3)(

23

++−= t t t

t X

v ( t )=dx

dt =t 2−5 t +6

!) v (t )=0=t 2

−5 t +6

t =3s∧ t =2 s

c)

5.62

.5

3)(

23

++−= t t t

t X


x ( t )=27

3 −45

2 +18+5=9.5m


Parat =0: x ( t )=5

Parat =4 s : x (t )=643−

80

2+24+5=10.3m

∴despla!amiento=∆ x=10.3−5=5.3m


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41

RESOLVER CADA CASO

1.

∫ dt t . ¿ t 2

2+c

2.

∫ dt t .2

*

t 3

3+c

3 .∫ t 3dt =

t 4

4+c

4. ∫ x4dx=

x5

5+c

5. ∫ x5

dx=

x6

6 +c

6. ∫ x "

dx= x

" +1

" +1+c

7. ∫ ( y+3 ) dy= y

2

2+3 y+c

8. ∫ ( y2− y+3 ) dy= y

3

3−

y2

2+3 y+c

9. ∫ ( x3+ x2+2 x+5 ) dx

¿ x

4

4+

x3

3+ x2+5 x+c

10.

x4

(¿+ x3+ x2+ x−1)dx

∫ ¿

¿ x5

5+ x

4

4+ x

3

3+ x

2

2− x+c

11. ∫k x "

dx k ϵ #

k ∫ x " dx k ϵ #

k ( x " +1) " +1

+C


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42

12. ∫2

4

x .dx=f (4 )−f (2)

x2

2 [42]= f (4 )−f (2 )=6

para x=4⇒(4)2

2=8

parax=2$ (2)2

2=2

13. ∫2

4

( x+1)dx

x2

2+ x[

42]=f ( 4 )− f (2 )=8

para x=4⇒(4)2

2+4=12

parax=2$ (2)2

2+2=4

14. ∫2

4

( x2+1)dx

x3

3+ x [42]=f ( 4 )−f (2 )=

76

3−

14

3=

62

3

parax=4⇒(4)3

3+4=

76

3

para x=2⇒(2)3

3+2=

14

3

15. ∫0

4

tdt

t 2

2 [40]=f (4 )−f (0 )=8−0=8

parat =4⇒(4)2

2=8

para t =0⇒ (0)2

2=0


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16. ∫0

3

(3 t −2 ) dt

3 t 2

2−2t [30]=f (3 )−f (0 )=18−0=18

para t =3⇒3(4)2

2−2 (3 )=24−6=18

parat =0⇒3 (0)2

2−2(0)=0

17. ∫0

1

(t 2+3 t −2)dt

t 3

3+3t 2−2 t [10]=f (1 )− f (0 )=

4

3−0=

4

3

parat =1⇒(1)3

3+3 (1)2−2(1)=

4

3

para t =0⇒(0)3

3+3(0)2−2(0)=0

18. ∫3

4

(t 2+3 t )dt

t 3

3+3t 2[43 ]= f (4 )−f (3 )=

208

3−36=

100

3

para t =4⇒(4)3

3+3(4 )2=

208

3

parat =3⇒

(3)3

3 +3(3)

2

=36

19. ∫0

1

(t 2−2)dt

t 3

3−2 t [10]= f (1 )−f (0 )=

5

3−0=

5

3

parat =1⇒(1)3

3−2(1)=

5

3


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parat =0⇒(0)3

3−2(0)=¿

20. ∫ %en (t ) dt

∫ %en (t )dt =−cost +C

21. ∫ %en (2t ) dt

u=2 t y du=2dt

eemplaFando

%en (u ) du

2=

1

2∫ senu .du=−1

2cosu+ & =¿−

1

2cos (2 t )+ &

∫ ¿

22. ∫ %en (3 t ) dt

u=3 t y du=3dt

eemplaFando

%en (u ) du

3=

1

3∫ senu.du=−1

3cosu+C =¿−

1

3cos (3 t )+C

∫ ¿

23. ∫ %en (4 t ) dt

u=4 t y du=4 dt

eemplaFando

%en (u) du

4=

1

4∫ senu .du=−1

4cosu+C =¿−

1

4cos (4 t )+C

∫¿

24. ∫ %en (5 t ) dt

u=5 t y du=5dt

eemplaFando

%en (u ) du

5=

1

5∫ senu.du=−1

5cosu+C =¿−

1

5cos (5 t )+C

∫ ¿

25. ∫cos (t ) dt


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45

∫cos ( t ) dt =sent +

26. ∫cos (2 t ) dt

u=2 t y du=2dt

eemplaFando

cos (u ) du

2=

1

2∫Cosu.du=1

2senu+C =¿

1

2sen (2 t )+C

∫ ¿

27. ∫cos (3 t ) dt

u=3 t y du=3dt

eemplaFando

cos (u ) du

3=

1

3∫Cosu.du=1

3%enu+C =¿

1

3%en (3 t )+C

∫ ¿

28. ∫cos ( 4 t ) dt

u=4 t y du=4 dt

eemplaFando

cos (u ) du

4=

1

4∫Cosu.du=1

4%enu+C =¿

1

4%en (4 t )+C

∫¿

29. ∫cos (5 t ) dt

u=5 t y du=5dt

eemplaFando

cos (u ) du

5=

1

5∫Cosu.du=1

5senu+C =¿

1

5sen (5 t )+C

∫ ¿

PROBLEMAS PROPUESTOS

+,IVEL I,TERMEDIO)


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1. na part


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47

a velocidad es 10 î (m .s−1 )

(5 )2−3 (5 )+C =10

C =0

a velocidad en el instante t =8s

(8 )2+3 (8 )=V (8s)

V (10 s)=88 î (m . s−1 )

3. na part


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48

∫ ( 4 t −5) dt +C =V ( t )

4∫ td t −5∫dt +C =V (t )

2t 2−5 t +C =V (t )

,n el instante t =3s 9 la velocidad es 20 î (m . s−1 )

2 (3 )2−5 (3 )+C =20

C =17


(10)2+5 (10 )+17=V (10 s)

V (10 s)=167 î (m .s−1 )

5. na part


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6. na part


50/84


50


(10)4−71=V (10 s)

V (10 s)=9929 î (m . s−1 )

8. na part


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−10 t +C =V (t )

,n el instante t =2 s 9 la velocidad es 20 î (m .s−1 )

−10 (2)+C =20

C =0


−10 (10)=V (10s )

V (10 s)=−100 î ( m.s−1 )

10. na part


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52

−5∫ tdt +35∫dt + = X (t )

−52

t 2+35 t +k = X (t )

,n el instante t =0s la posici;n es 10 î (m ) .

−52

t 2+35 t +k = X ( t )

−52

(0 )2+35(0)+k =10

k =10

Determine la posici;n en el instante t = $"s

−52

(10 )2+35 (10)+10= X (t )

X (t )=110 î (m ) .

11. na part


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C =60

Iallando la ecuaci;n de la posici;n

∫V (t )dt + = X (t )

∫ (−10 t +60 )dt +k = X (t )

−10∫ tdt +60∫dt +k = X (t )

−5 t 2+60 t +k = X ( t )


−5 t 2+60 t +k = X (t )

−5 (0 )2+60 (0)+k =10

k =10


−5

(10

)

2

+60

(10

)+10

= X (t )

X (t )=110 î (m ) .

12. na part


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54

,n el instante t =5s 9 la velocidad es 10 î (m .s−1 )

−52

(5)2+C =10

C =1452


∫V (t )dt + = X (t )

∫(−5 t 2+ 1452 )dt +k = X (t )

−5∫ t 2dt + 1452 ∫dt +k = X (t )

−53

t 3+

145

2t +k = X (t )


−53 t

3

+

145

2 t +k = X (t )

−53

(0 )3+ 1452

(0)+k =10

k =10


−53 (10 )

3

+145

2 (10)+10= X (t )

X (t )=−2795 î (m ) .

13. na part


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55

∫a (t )dt +C =V ( t )

∫ ( t −5 ) dt +C =V (t )

∫ tdt −5∫ dt +C =V (t )

t 2

2−5 t +C =V ( t )


(5)2

2

−5 (5 )+C =20

C =65

2


∫V (t )dt + = X (t )

∫(t 2

2−5 t +

65

2 )dt +k = X (t )

1

2∫ t 2 dt −5∫ tdt + 65

2 ∫ dt +k = X (t )

1

6t 3−

5

2t 2+

65

2t +k = X ( t )


1

6t 3−

5

2t 2+

65

2t +k = X (t )

1

6(0)3−

5

2(0 )2+ 65

2(0)+k =10

k =10



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56

1

6(10)3−

5

2(10 )2+

65

2(10)+10= X ( t )

X (t )=6085 î (m ) .

14. na part


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57

1

6t 3+

5

2t 2−

35

2t +k = X ( t )

1

6(0)3+

5

2(0 )2−

35

2(0)+ k =10

k =10


1

6(10)3+

5

2(10 )2−

35

2(10)+10= X ( t )

X (t )=5915 î (m ) .

15. na part


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58

1

3t 3+

t 2

2−10 t +k = X (t )


1

3t 3+

t 2

2−10 t +k = X (t )

1

3(0)3+

t 2

2−10(0)+k =10

k =10


1

3(10)3+

(10)2

2−10(10)+10= X (t )

X (t )=8803

î (m )

16. na part


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59


∫V (t )dt + = X (t )

∫(t

2

2 +t +5

2 )dt +k = X (t )

1

2∫ t 2 dt +∫ tdt + 5

2∫ dt +k = X (t )

1

6t 3+

t 2

2+5

2t +k = X (t )

,n el instante t =0

s la posici;n es10 ^

i (m ) .

1

6t 3+

t 2

2+5

2t +k = X ( t )

1

6(0)3+

t 2

2+5

2(0)+k =10

k =10


1

6(10)3+

(10)2

2+5

2(10)+10= X ( t )

X (t )=775

3î (m )

17. na part


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60

t 2

2+t +C =V ( t )


(5)2

2+(5 )+C =20

C =5

2


∫V (t )dt + = X (t )

∫(t 2

2+t +

5

2 )dt +k = X (t )

1

2∫ t 2 dt +∫ tdt + 5

2∫ dt +k = X (t )

1

6t 3+

t 2

2+5

2t +k = X (t )


1

6t 3+

t 2

2+5

2t +k = X ( t )

1

6(0)3+

t 2

2+5

2(0)+k =10

k =10


1

6(10)3+

(10)2

2+5

2(10)+10= X ( t )

X (t )=7753

î (m )

18. na part


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61

20 î (m .s−1 ) . ,n el instante t =0 s la posici;n es 10 î (m ) . Determine la

posici;n en el instante t =10 s

∫a (t )dt +C =V ( t )

∫ (2 t +3 )dt +C =V ( t )

2∫ tdt +3∫dt +C =V (t )

t 2+3 t +C =V (t )


(5)2+3 (5 )+C =20

C =−20


∫V (t )dt + = X (t )

∫ (t 2+3 t −20 ) dt +k = X ( t )

∫ t 2dt +3∫ tdt −20∫dt +k = X (t )

1

3t 3+

3 t 2

2−20 t +k = X (t )


1

3 t 3

+3 t

2

2 −20 t +k = X (t )

1

3(0)3+

3 t 2

2−20(0)+ k =10

k =10


1

3(10)3+ 3(10)

2

2−20 (10)+10= X ( t )


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62

X (t )=880

3î (m )

19. na part


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63

k =10


1

3(10)3−45(10)+10= X (t )

X (t )=−320

3î (m )

20. na part


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64

−13

t 3+80 t +k = X ( t )

−13

(0 )3+80( 0 )+k =1k =10


−13

(10)3+80 (10)+10= X (t )

X (t )=1430

3î (m )

21. na part


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65

−5 t 2+20 t +k = X ( t )


−5 t 2+20 t +k = X (t )

−5 (0 )2+20 (0)+k =10

k =10


−5 (10 )2+20(10)+10= X (t )

X (t )=−290 î (m ) .

22. na part


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66

−20∫ tdt +40∫ dt +k = X (t )

−10 t 2+40 t +k = X (t )

,n el instante t =2 s la posici;n es 10 î (m ) .

−10 t 2+40 t +k = X (t )

−10 (2 )2+40 (2 )+k =1k =−30


−10 (10)2+40 (10 )−30= X (t )

X (t )=−630 î (m ) .

23. na part


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67

∫V (t )dt + = X (t )

∫(−103 t 3−20

3 )dt +k = X ( t )

−103 ∫ t 3 dt −20

3 ∫ dt +k = X (t )

−56

t 4−

20

3t +k = X (t )


−56

t 4− 20

3t +k = X (t )

−56

(0 )4− 203(0)+k =10

k =10


−56

(10 )4−20

3(10)+10=10

X (t )=8390 î (m ) .

24. na part


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68

,n el instante t =4 s 9 la velocidad es −20 î (m. s−1 )

10 (4 )+C =−20

C =−60


∫V (t )dt + = X (t )

∫ (10 t −60) dt + = X ( t )

10∫ tdt −60∫ dt + = X (t )

5 t 2−60t +k = X (t )


5 t 2−60t +k = X (t )

5 (0 )2−60(0)+k =10

k =10


5 (10 )2−60(10)+10= X ( t )

X (t )=−90 î (m ) .

25. Determinar el camBio de la velocidad: (V =∫2

4

tdt

t 2

2 [42]= f (4 )−f (2)=6 m.s−1

para t =4⇒(4)2

2=8

parat =2⇒

(2)2

2 =2


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69

26. Determinar el camBio de la velocidad: (V =∫2

4

(t +1)dt

t 2

2+t

[4

2

]=f (4 )− f (2 )=8m .s−1

para t =4⇒(4)2

2+4=12

parat =2⇒(2)2

2+2=4

56. Determinar el camBio de la velocidad:( )∫ −=∆

6

2

2 2 dt t V

∫2

4

(t 2−2 ) . dt =[ t 3

3−2 t ]

2

4

=6

3

3−2.6−(2

3

3−2.2)

) ¿60−(−1,3 )=61,3m.s−1

57. Determinar el camBio de la velocidad:

( )∫ −=∆

4

3

2 1 dt t V

(t 2−1 ) . dt =¿[ t 3

3−t ]

3

4

=43

3−4−(3

3

3−3)

∫3

4

¿

)=17,3−(6 )=11,3 m. s−1


( )∫ =∆4

0

.2 dt t V

∫0

4

(2 t ) . dt =[ t 2 ]04

=42−02

) ¿16m.s−1


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70


( )∫ −=∆3

0

23 dt t V

∫0

3

(3 t −2 ) . dt =[3 t

2

2 −2t ]03

=3 .32

2 −2.3−(3 .0

2

2 −2.0))=7,5−0=7,5m .s−1


( )∫ =∆3

0

23 dt t V

∫0

3

(3 t 2 ). dt ¿=[t 3

]0

3

=33−03

) ¿27m.s−1


∫ =∆3

0

)( dt t SenV

sin ( t ) . dt =[−cost ]03=¿−cos3−(−cos0)

∫0

3

¿

) ¿− (−0,98 )−(−1 )=1,98m.s−1


∫ =∆3

0

)( dt t CosV

cos ( t ) . dt =¿ [sin t ]03=sin3−(sin0 )

∫0

3

¿

) ¿0,14−(0 )=0,14m. s−1

63. Determinar el desplaFamiento:

( )3

2

1

3 2 X t dt ∆ = −∫

∫1

3

(3 t 2−2 ) .dt =[ t 3−2t ]13


71/84


71

¿33−2.3−(13−2.1 )

)=21− (−1 )=22m .s−1


( )∫ +=∆4

2

53 dt t X

∫2

4

(3 t +5 ) . dt =[3 t 2

2+5 t ]

2

4

=3.4

2

2+5.4−(3.2

2

2+5.2)

)=44− (16 )=28m .s−1


( )∫ +=∆5

2

2 10.3 dt t X

∫2

5

(3 t 2+10 ) . dt = [t 3+10 t ]25

=53+10.5−(23+10.2 )

)=175−28=147m .s−1


( )∫ +=∆6

2

2 10.3 dt t X

∫2

6

(3. t 2+10 ). dt =[ t 3+10 t ]26

=63+10.6− (23+10.2 )

)=276−28=248m.s−1


( )∫ −=∆9

322 dt t X

∫3

9

(2 t −2) . dt =[ t 2−2 t ]39

=92−2.9−(32−2.3 )

) ¿63−3=60m. s−1


∫ =∆9

3

)( dt t Sen X


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72

∫3

9

sin ( t ). dt =[−cos t ]39=−cos9 * cos3

)=−(−0,91 )−(−0,98 )=1,89m .s−1


∫ =∆8

3

)2( dt t Sen X

sin (2t ) . dt =¿[−cos2 t 2 ]38

=−cos16

2−(−cos62 )

∫3

8

¿

) ¿− (−0,47 )−(−0,48 )=0,95m .s−1


∫ =∆9

3

)( dtost Cos X

cos ( t ) . dt =¿ [sin t ]39=sin9− (sin3 )

∫3

9

¿

) ¿0,41−(0,14 )=0,27 m.s−1


∫ =∆8

3

)2( dt t Cos X

cos (2 t ) . dt =¿[ sin 2t 2 ]38

=sin 16

2−( sin 62 )

∫3

8

¿

)=−0.14−(−0,13 )=−0,01m. s−1

P4*,3A6 P4P,646

1. na part


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73

m.s-2. Para0=t

la rapideF esV

8 para4=t s

la rapideF es3V

. Determinar la

rapideF para6=t s

t =0s ) v0=V t =6 s )v

6=+

t =4 s) v4=3V

v ( t )=∫at . dt +C 1

v ( t )=t 2

2+8 t +C

1

) v (0)=C 1=V 8 v (4 )=8+32+v=3V

40=2V 20=V

) v ( 6)=18+48+20=86m. s−1

2. na part


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74

3. na part


75/84


75

5. na part


76/84


76

$) x (2)=8−8+5=5 i

x (1)=2 i

#=5 i+2i=7 i

2) x (4 )=48+5=53

x (2)=8−8+5=5

-=53−2=51

8. 6i la le8 de movimiento de una part


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77

10. ,l movimiento de una partnido por la relaci;n( ) it t t x ˆ61562 23 −+−=

9 donde ?@ se e@presa en metros 8 ?t en segundos.Determine:$) la posici;n de la part


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78

X.a) CDespus de cu7ntos segundos se encuentran untos por segunda veFEB) C,n u posici;n se encuentran cada uno de los m;viles cuando sedetienen instant7neamenteEc) Determine la distancia de m7@imo aleamiento entre los m;viles.

a) −8t +2 t 2=10 t −t 2

t (3 t −18 )=0

)t =0s 4 3 t −18=0

t =6s - por segunda veFM

B) x1=−8t +2 t 2

x2=10 t −t 2

) v1=

4 t −8 )v

2=−2t +10

0=4 (t −2) 0=2(−t +5 )

t =2 s

t =5 s

) x2=−16+8 )x

5=50−25

) x2=−8m )x

5=−25m

13. n m;vil se mueve a lo largo del ee ?@ d acuerdo a la le8:

22 2 3 x

V t t = + + donde

xV

se mide en mKs 8 ?t en segundos. 6aBiendo ue cuando0t =

la posici;n es1 x m=

a) ,ncontrar el valor de ?@ cuando2t s=

B) Determine la aceleraci;n cuando3t s=

x (t )=∫v ( t ) . dt +C 2) x (t )=t 3+ t 2+2 t +1

at =dv

dt ) at =6 t

2+¿

a) ,n t =2s

x (2)=8+4+4+1=17

B) nt =3 s

a(3)=6.3+2=20


79/84


79

14. na part


80/84

#

X

4t (s)

-#

#

(*)

X (m)

$2

(A)

'

#

4

t (s)

Para el proBlema #

(*)

V (mKs)

2

2

(A)


80

6i el m;vil parte del reposo en la posici;n3 x m= +

9 determinar la velocidad

cuando pasa por0 x =.

at

=dv

dt ) v

x=−2 x2+C

v0)x=3

−2 x2+C =0)−2.32+c=0 )C =18

v t =−2 x2+C +uando x=0

v t =C )v t =18m.s−1

2. a gra>ca x t −

muestra la le8 de movimiento de los m;viles A 8 * ue semueven rectilco se oBserva ue la ma8or distancia de separaci;n se dar7 en eltiempo t= s

) por teorema de riangulo recto saBemos ue x2=9.3)x=3 √ 3 8 esta es la ma8or

distancia

3. a gra>cav t −

mostrada descriBe el movimiento de dos m;viles A 8 * ue semueven son el ee ?@. 6i los dos m;viles parten desde un mismo punto en

0t =9 Hallar la ventaa ue le sac; el m;vil A al

m;vil * Hasta el instante en ue susaceleraciones se Hacen iguales. ,n la >gura lascurvas son cuartos de circunferencias.


81/84

4

t (s)

Para el proBlema !

V (mKs)

2

2


81

da=π 32

2

4=π

d4=4−π

) da−d4

π −(4−π )=π −4+π =2 π −4

4. a gra>ca muestra la manera como var


82/84


82

5. as le8es del movimiento de dos part


83/84


83

7. as le8es del movimiento de dos part


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Profesor de Física WALTER PEREZ TERREL TEMA: Cinemáticak =3

B v t =t 3−32

⃗v=¿⃗ x=∫ t 3

−32=

t 4

4 −32t +C 2

∫¿

Para t =4 Posici;n x=0

44

4−3234+C

2=0 C 2=64

⃗ x=

t 4

4 −32t +64

ejercicios fisica cinematica

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