A

V t

A t 0 V t a v v

?t

A a

V v

0 v

if

f

i

T

V a

T a V 0 t a v v

? a

T t

0 v

V v

if

f

i

Problem 1: A commuter backs her car out of her garage with a constant acceleration of A. Assume that her initial motion is in the positive direction. Part (a) How long does it take her to reach a speed of V in seconds? Part (b) If she then brakes to a stop in T seconds, what is her acceleration in meters per square second?  

Solution:

(a)

 

 

 

 

(b)   

 

 

 

 

 

 

 

 

 

 

 

+ A 

 

Problem 2: A car begins at rest at t0 = 0. The car starts moving and finally covers a distance d = D in a time tf = T. In a coordinate system with north being the positive x-direction, the car's motion is in the southern direction (see figure). Part (a) What was the car's average speed during this period, in meters per second? Part (b) What was the car's displacement in the x-direction Δx, in meters, during this period? Part (c) What was the car's average velocity vavg, in meters per second, during this period? Part (d) If the car's final velocity at tf was vf = -V, what was the car's average acceleration aavg, in meters per square second, during this period? 

Solution:

(a)

case)present in trun - UNo ( T

D

t

distance speed Average

(b)

) than xnegative more is x(Note D-

x xx

if

if

(c)

) (b)part from -Dx( T

D

t

x velocity Average

(d)

+ (North) D 

T

V- a aT 0 V- at vv

?a

T t

V v

0v

if

f

i

Problem 3: A student begins at rest and then walks north with an average speed of v1 = V1. The student then turns south and walks with an average speed v2 = V1. Part (a) What was the student's overall average velocity vavg, x, in m/s, for the trip assuming the student spent equal times at speeds v1 and v2? Part (b) If the student traveled for T1seconds at speed v1, and for T2 seconds at speed v2, what was the net displacement d (in meters) during the trip? Part (c) Assuming that it took the student t = T’ to reach the average speed v1 from rest, what was the student's average acceleration magnitude, aavg in m/s2, during this time?

Solution:

(a) Assume the student starts walking towards north from the origin for a time t.

Coordinate of student when she turns around = V1t

Coordinate of student when she complete walking = V1t – V2t = (V1 – V2) t

2

VV

2t

t)V(V

t

xv

t )V(V 0 - t )V(V x

2121 x avg,

2121

(b)

Coordinate of student when she turns around = V1 T1

Coordinate of student when she complete walking = V1t – V2 T1

2211 TVTV x d

(c)

T'

V

T'

0V

t

va 11

x avg,

+ (North) V1 

O 

2BA 1))((2B A vs,0.1At t

2Bt A ) tB +A t (dt

d

dt

dv a

tB +A t = v(t)

1

2

2

3

26B +4A

-2B)(A )1-(3 3

B + )1-(3

2

A

t3

B +

2

tA

dt ) tB +(A t dt v(t) =x

3322

3

1

32

23

1

3

1

Problem 4: A particle’s velocity along the x-axis is described by v(t) = A t + B t2,

where t is in seconds, v is in meters per second. Part (a) What is the acceleration, in meters per second squared, of the particle at t0 = 1.0 s? Part (b) What is the displacement, in meters, of the particle between t0 = 1.0 s and t1 = 3.0 s? Part (c) What is the distance traveled, in meters, by the particle between t0 = 1.0 s and t1 = 3.0 s? Solution:

(a)

(b) 

(c)

6B

A9B+

2

9A

3

B

2

A

6B

Add is 3s t and 1sbetween t traveleddistance Total

6B

A9B+

2

9A

3B

A -

2B

A9B+

2

9A

B

A-

3

B +

B

A-

2

A3

3

B + 3

2

A

t3

B + t

2

A

dt ) tB +(A t dt v(t) =d

3s tand B

A-Between t

3

B

2

A

6B

A

3

B +

2

A

3B

A -

2B

A

1 3

B + 1

2

A

B

A-

3

B +

B

A-

2

A

t3

B + t

2

A

dt ) tB +(A t dt v(t) =d

B

A- t and 1s Between t

B

A- t 0 tB +A t when 0, at v around turnsparticle that theNote

33

21

3

33

3232

3

B

A-

32

23

B

A-

3

B

A-

2

3

33

3232

B

A-

1

32

2B

A-

1

B

A-

1

1

2

ms

1

m

1

s

mB of Unit

s

m m B of Unit

22

B

A - x or 0 x

0 Bx) x(A

0 xB +A x 0 = v(t)

motion. ofdirection it chages particle when the0v

xB +A x = v(t)

2

2

2B)A)(B +A ( a

1m At x

2Bx)A)( xB +A x (dx

dv v a

xB +A x = v(t)

2

2

Problem 5: A particle’s velocity along the x-axis is described by v(x) = A x + B x2,

where x is in meters, v is in meters per second. Part (a) What are the correct units for B? Part (b) At what position, in meters, other than the origin, will the particle change its direction of motion? Part (c) What is the acceleration, in meters per square second, of the particle at x1 = 1.0 m? Solution: (a) Unit for Bx2 should be the same as that of v, so Bx2 should have unit m/s. (b)  

(c) 

Problem 6: A bicycle initially at rest accelerates for t = T seconds at a rate of a = A m/s2. Part (a) Solve numerically for the distance, d, the bicycle traveled during this period in meters. Part (b) Write an expression for the final speed, v, of the bicycle after time t. Part (c) Solve numerically for the final speed v, in m/s.

Solution:

(a)

2

2f

2f

2iif

f

i

i

AT2

1 |x| d turn,- Uno is thereSince

AT2

1 x

AT2

1 T 00 x at

2

1 t vxx

? x

0v

T t

Aa

0x

(b) and (c)

(c))(part AT v

AT0 v

(b))(part at v v

? v

0v

T t

Aa

0x

f

f

if

f

i

i

+ A 

O 

AT60V v

A(60T)V v

at v v

? v

Vv

s 60T Tmin t

Aa

0x

if

if

if

f

ii

i

A'

AT60V t

tA' AT60V0

at v v

? t

stop) to(come 0 v

(a))part (from AT60Vv

ng)deceleratibecause(negstive A'a

i

i

if

f

ii

1800AT TV60 |x| d turn,- Uno is thereSince

1800AT TV60 x xx

1800AT TV60 x

A(60T)2

1 T 60V0 x at

2

1 t vxx

? x

Vv

s 60T Tmin t

Aa

0x

2i

2iif

2if

2if

2iif

f

ii

i

Problem 7: Freight trains can produce only relatively small accelerations and decelerations. Part (a) What is the final velocity, in meters per second, of a freight train that accelerates at a rate of A m/s2 for T min, starting with an initial velocity Vi m/s? Part (b) If the train can slow down at a rate of A’ m/s2, how long will it take to come to a stop from this velocity in seconds? Part (c) How far will it travel in part (a), in meters? Part (d) How far will it travel in part (b) in meters?

Solution:

(a)

(b)

(c)

+ Vi 

O 

Nose of freight train 

2A'

AT)60(V |x| d turn,- Uno is thereSince

2A'

AT)60(V x xx

2A'

AT)60(Vx x

A'

AT)60(V

2

1

A'

AT)60(Vx x

) A'

AT60V(A'

2

1 )

A'

AT60VAT)(60(Vx x at

2

1 t vxx

? x

stop) to(come 0v

(b))part (from A'

AT60V t

A'a

(a))part of (v AT60Vv

(c))part of (x 1800AT TV60x

2i

2i

if

2i

if

2i

2i

if

2iiiif

2iif

f

f

i

fii

f2

ii

(d)

A

AL2VV t

2A

AL84V2V t

2A

A(2L)4(-2V)(-2V)- t

0 2LVt 2At

At2

1 t V0L at

2

1 t vxx

? t

Vv

Lx

0)(A A a

0x

2

2

2

2

22iif

i

f

i

Problem 8: An express train passes through a station which is L m long. It enters with an initial velocity of V m/s and decelerates at a rate of A m/s2 as it goes through. Part (a) How long, in seconds, is the front of the train in the station? Part (b) How fast is it going when the nose leaves the station in m/s? Part (c) If the train is B m long, how long, in seconds, after the front of the train enters the station does the end of the train leave the station? Part (d) What is the velocity of the end of the train as it leaves in m/s?

Solution:

(a)

We have to assume V2-2AL 0. If this is not satisfied, there will be no (real) solution and this corresponds to the case that the station is too long and the train does not have enough speed to reach the other end of the station.

If V2-2AL > 0, note that V AL2V2 so there will be two solutions and both are positive! Which one is the answer? We should choose the smaller value as the answer because the large value corresponds to the time after the train overshoots and returns to that point again the second time. So it will take a

timeA

AL2VV t

2 for the front of the train to pass through the station.

x=0  x=L

station

Block dot represents front of train

AL2V v

)AL2V(V-V v

)A

AL2VVA(V v

at v v

?vA

AL2VV t

Vv

Lx

0)(A A a

0x

2f

2f

2

f

if

f

2

i

f

i

A

B)A(L2VV t

2A

B)A(L84V2V t

2A

B)]A[2(L4(-2V)(-2V)- t

0 B)2(LVt 2At

At2

1 t V0L at

2

1 t vxx

? t

Vv

BLx

0)(A A a

0x

2

2

2

2

22iif

i

f

i

(b)

(c) When the end of train leaves the station:

x=0  x=L

stationBlock dot represents front of train

x=L+B 

End of train is here

B)A(L2V v

)B)A(L2V(V-V v

)A

B)A(L2VVA(V v

at v v

?vA

A(LB)2VV t

Vv

BLx

0)(A A a

0x

2f

2f

2

f

if

f

2

i

f

i

Using the same argument in part (a), the larger value is for the overshoot case, so it will take

A

B)A(L2VV t

2 for the train to leave the station completely. 

 

(d) Note that the front and end of the train always have the same velocity, even though the velocity is 

changing with time.

6

tg v

tg/6)(0 v

at v v

?v

rest) from (released 0v

part) for this need (no 0y

0)m/s 9.8(g g/6a

d y

f

f

if

f

i

f

2

i

6

T g vf

12

gT d

T g/62

1 T 0 d 0 at

2

1 t vyy

?d y

Tt

part) for this need no (a),part fromresult (new 6

gTv

rest) from (released 0v

0y

0)m/s 9.8(g g/6a

2

22iif

i

f

i

f

2

Problem 9: Gravitational acceleration on the moon is one sixth of that on Earth. A ball released from rest above the surface falls from height d in a time of t = T seconds. Part (a) Write an expression for the final velocity vf of the ball when it impacts the surface of the moon assuming it is dropped from rest. This expression should be in terms of only, g (the gravitational acceleration on Earth), and time, t Part (b) Calculate the final velocity, vf, numerically in m/s. Part (c) Calculate the height, d (in meters), from which the ball was dropped. Part (d) From how high up would this ball need to be dropped on the earth, de (in meters), if it took the same time to reach the ground as it did on the moon? Solution: (a) (b) (c)

y

g 

2

gT y

Tg2

1 T 0 y 0 at

2

1 t vyy

? y

Tt

rest) from (released 0v

0y

0)m/s 9.8(g ga

2

i

2i

2iif

i

i

f

2

(d)

g

gH2VV t

2g

gH84V2V t

2g

g(-2H)4(-2V)(-2V)- t

0 2HVt 2gt

gt2

1 t VH0 gt

2

1 Vt yy

? t

Vv

0y

0)(g -ga

Hy

2

2

2

2

22if

i

f

i

Problem 10: A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of V m/s, and her takeoff point is H m above the pool. Part (a) How long are her feet in the air, in seconds? Part (b) What is her highest point above the board, in meters? Part (c) What is her velocity when her feet hit the water, in meters per second? Assume the vertical component of the velocity is positive upwards. Solution:

Note that V AL2V2 so there will be two solutions with one positive and one negative. It is natural to choose the positive answer, and that is indeed the right choice. Why there is a negative answer and what does it mean? If we extend the dotted line in the figure backward in time, we can see the diver’s toe was once at the water lever when t<0 and that’s the negative answer represents.

Diver’s feet will be in air for a time g

gH2VV t

2 .

water 

y 

g 

i 

V 

Black dot represents toe of diver 

y=0 f 

2g

VHy is board theabovepoint highest Her

2g

VHy

2g

V

g

VHy

)g

Vg(

2

1 )

g

V( VHy at

2

1 t vyy

:now t know wesince problem no is There .for v askingactually is problem But the

g

V t

gtV0

at v v

?t

0v

Vv

0)(g ga

Hy

2

f

2

f

22

f

2f

2iif

f

if

f

i

i

gH2V- v

)gH2V(V-V v

)g

gH2VV(-g)(V v

at V vat vv

?v

(a))part from(result g

gH2VV t

Vv

0y

0)(g -ga

Hy

2f

2f

2

f

fif

f

2

i

f

i

(b)

(c)

y 

g V 

i 

Black dot represents toe of diver 

water 

f 

y=0 

water 

y 

g 

i 

V 

Black dot represents toe of diver 

y=0 f 

VT - gT2

1 H

gT2

1 T V0H- at

2

1 t vyy

? -Hy

Tt

Vv

0)(g ga

0y

2

22iif

f

i

i

Problem 11: A rock takes T s to hit the ground when it is thrown straight up from the cliff with an initial velocity of V m/s. Part (a) Calculate the height of the cliff in m. Part (b) How long would it take to reach the ground if it is thrown straight down with the same speed? Solution: Let the height of the cliff be H. (b)

water 

y 

g 

i 

V 

Black dot represents rock 

y=‐H 

y=0 

f 

water 

y 

g 

i 

V Black dot represents rock 

y=‐H 

y=0 

f 

g

)gT -(2VTVV- t

2g

)gT -(2VT4(2V)2V- t

0)gT -(2VTVt 2 tg

tg t V2gT -2VT

(-g)t2

1 t V)(0gT

2

1 -VT at

2

1 t vyy

?t

part(a)) (from gT2

1 -VT -Hy

Vv

0)(g ga

0y

22

22

22

22

222iif

2f

i

i

g

)gT-Vg2(VV- t

22

Note that according to part (a), gT2 -2VT=2H > 0, so there are two roots for t, one positive and one negative.  If we extend the dotted line in the figure backward in time, we can see the diver’s toe was once at the water lever when t<0 and that’s the negative answer represents. We will take the positive root as the answer. It will take for the rock to reach the bottom of the cliff.

meter)(in g2

1

0-g2

1 |vv| object by the traveleddistance

between,in turn - Uno is There

g2

1 y

g(1)2

1 T 00y at

2

1 t vyy

? y

second)(first 1t

0v

0)(g ga

0y

if

f

2f

2iif

f

i

i

Problem 12: An object is dropped from a height of H m above ground level. Part (a) Determine the distance traveled during the first second in meters. Part (b) Determine the final speed at which the object hits the ground in m/s. Part (c) Determine the distance traveled during the last second of motion before hitting the ground in meters. Solution: (a) (b)

water 

y 

g 

i 

Black dot represents object 

y=‐H 

y=0 

f 

water 

y 

g 

i 

V Black dot represents rock 

y=‐H 

y=0 

f 

2gH |v| Speed

2gH v)g

2Hg)((0 vat vv

:now t know we

since problem no is There ground. thehitsobject when thespeed for the asking is problem But the

g

2H t

g

2H t

gt2

1 T 00H- at

2

1 t vyy

?t

H- y

0v

0)(g ga

0y

f

ffif

2

22iif

f

i

i

?y

ground) hitting before second (1 1 - g

2Ht

0v

0)(g ga

0y

f

i

i

We will first determine the time when the object hits the ground. (c)

water 

y 

g 

i 

V Black dot represents rock 

y=‐H 

y=0 

f 

meter)(in g2

12gH

g)2

12gH-(H- H

1)g

2H2-

g

2Hg(

2

1- H

1)-g

2Hg(

2

1- H secondlast in theobject by the traveleddistance

1)-g

2Hg(

2

1 y

1)-g

2H(-g)(

2

1t 0 0 y at

2

1t v y y

2

2f

2f

2yif

g

gH2v- v t up, way on the isit becauseroot smaller take

g

gH2vv t

g2

gH8v42v t

0H2 t2v- gt

gt2

1 t vH at

2

1 t v yy

? tt

H y

g- a

0 y

2ii

1

2ii

1

2ii

1

1i2

1

211i

2iif

1

f

i

g

L)g(H2vv t down, way on the isit becauseroot larger take

g

L)g(H2vv t

g2

L)g(H8v42v t

0L)H(2 t2v- gt

gt2

1 t vLH at

2

1 t v yy

? tt

LH y

g- a

0 y

2ii

2

2ii

2

2ii

2

2i2

2

222i

2iif

2

f

i

Problem 13: A ball is thrown straight up from ground level. It passes a L-m-high window. The bottom of the window is H m off the ground. The time that elapses from when the ball passes the bottom of the window on the way up, to when it passes the top of the window on the way back down is T s. Part (a) What was the ball’s initial speed in m/s? Solution: First calculate the time when it first passes the bottom of the window on the way up, in terms of the unknown initial velocity vi: Next calculate the time when it passes the top of the window on the way back down, in terms of the unknown initial velocity vi:

window

y

y = H

y = L+H

vi

also. good isit so

root, square after the s

m become unit will this,

s

m m

s

m isunit its and gH2 is term thirdThe

good. isit so root, square then and

squareafter same remain the unit will this,s

m isunit its and

T

L

2gT

2gL is termsecond The

good. isit so root, square then and

squareafter same remain the unit will this,s

m s

s

m isunit its and

2

gT

2gT

Tg is first term The

answer. on thecheck dimension a have toidea good a isit this,like problem dcomplicateFor

gH2 2gT

gL2Tg v

gH2 p v

gH2 p v gH2 v pBut

2gT

gL2Tg p

gL2Tg 2gTp

2gTp - Tg gL2

p 2gTp - Tg gL2p

) p - (gT gL2p

p - gT gL2p gT p gL2p

: toreducesequation Above

p. know weif solved is that vNote

gL2pL)g(H2 vso and gH2 v plet signs,root square fnumber reduce To

gT gH2v L)g(H2v

gT ]gH2v[v] L)g(H2v[v

T g

gH2vv

g

L)g(H2vv T t- t

2

2

2

2

22

222

i

2i

22i

2i

2

22

22

22

2222

22

22

i

22i

2i

2

2i

2i

2ii

2ii

2ii

2ii

12

Now set up equations:

ground). (above 2g

V H iscoin by the reachedheight maximum The

2g

V H x

g

V

2

1

g

V H x

g

V(-g)

2

1

g

VV H x at

2

1 t v xx

? x,g

V At t

g

V t

gt - V 0 at vv

?t

point.highest reach the toneeded timeout the figureFirst

0 vpoint,highest At

g- a

V v

Hx

2

2

f

22

f

2

f2

iif

f

if

f

i

i

gT 2

1 VT H x

(-g)T 2

1 VT H x at

2

1 t v xx

?x

Tt

g- a

V v

Hx

2f

2f

2iif

f

i

i

Problem 14: A coin is dropped from a hot-air balloon that is H m above the ground and rising at V m/s upward. For this problem use a coordinate system in which up is positive. Part (a) Find the maximum height reached for the coin in meters. Part (b) Find its position T s after being released in m. Part (c) Find its velocity T s after being released m/s. Part (d) Find the time before it hits the ground in seconds. Solution:

(b)

y 

y=H 

y=0 

V 

T g- V v

T (-g) V v ta vv

?v

Tt

g- a

V v

Hx

f

fif

f

i

i

g

gH2V V t

g2

gH84V 2V t

g2

g)(-2H)(4(-2V) 2V t

0 2H -2Vt - gt

(-g)t 2

1 Vt H 0 at

2

1 t v xx

?t

0x

g- a

V v

Hx

2

2

2

2

22iif

f

i

i

g

gH2V V t

2

(c) (d) There are two roots, one positive and one negative. If we extrapolate the path of the coin backward before r=0, we will find a time when the coin rising from the ground. This correspond to the negative answer and this is not the answer we want. So we choose the positive root and it will take a time of For the coin to fall back to ground.