physics 1443

akamo (qaa67) Chapter 2 Spivey (81466) 1

This print-out should have 19 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.The assignment is due Tuesday, Sept 7 at

10pm. The solutions will automatically beposted 10 minutes later. Remember: 1) Dontguess on the multiple choice! 2) Dont writethe units (like meters) in the entry field, justthe numerical values.

001 (part 1 of 2) 10.0 pointsA person travels by car from one city toanother with different constant speeds be-tween pair of cities. She drives for 33.6 minat 77.8 km/h, 13.2 min at 99.7 km/h, and57.4 min at 40.5 km/h, and spends 21.2 mineating lunch and buying gas.Find the distance between the initial and

final cities along this route.

Correct answer: 104.247 km.

Explanation:

Let : t1 = 33.6 min ,

v1 = 77.8 km/h ,

t2 = 13.2 min ,

v2 = 99.7 km/h ,

t3 = 57.4 min ,

v3 = 40.5 km/h ,

t4 = 21.2 min , and

v4 = 0 km/h .

x = vavg t

x1 = (77.8 km/h)(33.6 min)

(1 h

60 min

)

= 43.568 km ,

x2 = (99.7 km/h)(13.2 min)

(1 h

60 min

)

= 21.934 km , and

x3 = (40.5 km/h)(57.4 min)

(1 h

60 min

)

= 38.745 km ,

so the total distance traveled is

xtot = x1 +x2 +x3

= 43.568 km+ 21.934 km+ 38.745 km

= 104.247 km .

002 (part 2 of 2) 10.0 pointsFind the average speed for the trip.

Correct answer: 49.8789 km/h.

Explanation:

ttot = t1 +t2 +t3 +t4

= 33.6 min + 13.2 min

+ 57.4 min + 21.2 min

= 125.4 min ,

so the average velocity is

vavg =xtotttot

=104.247 km

125.4 min

(60 min

1 h

)

= 49.8789 km/h .

003 10.0 pointsA rocket initially at rest accelerates at a rateof 70 m/s2 for 0.84 min.What is its speed at the end of this time?

Correct answer: 3528 m/s.

Explanation:

Let : a = 70 m/s2 and

t = 0.84 min .

vf = vo + a t = a t

= (70 m/s2) (0.84 min)60 s

1 min

= 3528 m/s .

004 (part 1 of 3) 10.0 points


A particle moving uniformly along the x axisis located at 9.71 m at 1.31 s and at 4.89 m at3.95 s.Find its displacement during this time in-

terval.

Correct answer: 4.82 m.Explanation:

Let : xf = 4.89 m and

xi = 9.71 m .

The displacement is

x = xfxi = 4.89 m9.71 m = 4.82 m .

005 (part 2 of 3) 10.0 pointsWhat is its average velocity during this timeinterval?

Correct answer: 1.82576 m/s.Explanation:

Let : ti = 1.31 s and

tf = 3.95 s .

The average velocity is

v =x

t=

xf xitf ti

=4.89 m 9.71 m3.95 s 1.31 s = 1.82576 m/s .

006 (part 3 of 3) 10.0 pointsCalculate the particles average speed duringthis time interval.

Correct answer: 1.82576 m/s.

Explanation:

The average speed is the magnitude of theaverage velocity:

speedavg = |v| = |1.82576m/s| = 1.82576 m/s .


Consider a toy car which can move to theright (positive direction) or left on a horizon-tal surface along a straight line.

car

v

O +

What is the acceleration-time graph if thecar moves toward the right (away from theorigin), speeding up at a steady rate?

1. None of these graphs is correct.

2. t

a

3. t

a

4. t

a

5. t

a

correct


6. t

a

7. t

a

8. t

a

Explanation:

Since the car speeds up at a steady rate,the acceleration is a constant.

008 (part 2 of 5) 10.0 pointsWhat is the acceleration-time graph if the carmoves toward the right, slowing down at asteady rate?

1. t

a

correct

2. t

a

3. t

a


5. t

a

6. t

a

7. t

a

8. t

a

Explanation:

Since the car slows down, the accelerationis in the opposite direction.

009 (part 3 of 5) 10.0 pointsWhat is the acceleration-time graph if the carmoves towards the left (toward the origin) ata constant velocity?

1. t

a


3. t

a


4. t

a

correct

5. t

a

6. t

a

7. t

a

8. t

a

Explanation:

Since the car moves at a constant velocity,the acceleration is zero.

010 (part 4 of 5) 10.0 pointsWhat is the acceleration-time graph if the carmoves toward the left, speeding up at a steadyrate?

1. t

a

2. t

a


4. t

a

correct

5. t

a

6. t

a

7. t

a

8. t

a

Explanation:

The same reason as Part 1.

011 (part 5 of 5) 10.0 pointsWhat is the acceleration-time graph if the carmoves toward the right at a constant velocity?


1. t

a

2. t

a


4. t

a

5. t

a

6. t

a

7. t

a

correct

8. t

a

Explanation:

The same reason as Part 3.

012 10.0 pointsA car traveling in a straight line has a velocityof 3.07 m/s at some instant. After 7.53 s, itsvelocity is 11.3 m/s.What is its average acceleration in this time

interval?

Correct answer: 1.09296 m/s2.

Explanation:

Let : v1 = 3.07 m/s ,

t = 7.53 s , and

v2 = 11.3 m/s .

The average acceleration is

aav =v

t=

vf vit

=11.3 m/s 3.07 m/s

7.53 s

= 1.09296 m/s2 .

013 10.0 pointsWhen the shuttle bus comes to a sudden stopto avoid hitting a dog, it decelerates uniformlyat 4.0 m/s2 as it slows from 9.5 m/s to 0 m/s.Find the time interval of acceleration for

the bus.

Correct answer: 2.375 s.

Explanation:

Let : aavg = 4.0 m/s2 ,vi = 9.5 m/s , and

vf = 0 m/s .

aavg =v

t=

vf v0t

=v0t

t =viaavg

=9.5 m/s4 m/s2

= 2.375 s .



A plane cruising at 260 m/s accelerates at25 m/s2 for 4.3 s.What is its final velocity?

Correct answer: 367.5 m/s.

Explanation:

Let : v = 260 m/s ,

a = 25 m/s2 , and

t = 4.3 s .

v = v0 + a t

= 260 m/s + (25 m/s2) (4.3 s)

= 367.5 m/s .

015 (part 2 of 2) 10.0 pointsHow far will it have traveled in that time?

Correct answer: 1349.12 m.

Explanation:

sf = so + v0 t+1

2a t2

= v t+1

2a t2

= (260 m/s) (4.3 s) +1

2(25 m/s2) (4.3 s)2

= 1349.12 m .

016 10.0 pointsThe barrel of a rifle has a length of 0.974 m. Abullet leaves the muzzle of a rifle with a speedof 635 m/s.What is the acceleration of the bullet while

in the barrel? A bullet in a rifle barrel doesnot have constant acceleration, but constantacceleration is to be assumed for this problem.

Correct answer: 2.06994 105 m/s2.Explanation:

Let : = 0.974 m ,

vi = 0 m/s , and

vf = 635 m/s .

Under constant acceleration,

v2f = v2

i + 2 a = 2 a

a =v2f2

=(635 m/s)2

2 (0.974 m)

= 2.06994 105 m/s2 .

017 (part 1 of 2) 10.0 pointsA ball is thrown upward with an initial verti-cal speed of v0 to a maximum height of hmax.

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

v 0

hm

ax

What is its maximum height? The acceler-ation of gravity is g . Neglect air resistance.

1. hmax =v20

g

2. hmax =

3 v2

0

22 g

3. hmax =

5 v2

0

22 g

4. hmax =

3 v2

0

2 g

5. hmax =3 v2

0

4 g

6. hmax =v202 g

7. hmax =v20

2 gcorrect


8. hmax =v20

4 g

9. hmax =5 v2

0

8 g

Explanation:

With up positive, a = g and for the up-ward motion tf = 0, so for constant accelera-tion,

v2f = v2

0 + 2 a (x x0)0 = v20 2 g (hmax 0)

hmax =v20

2 g.

018 (part 2 of 2) 10.0 pointsFind the speed of the ball (in terms of theinitial speed v0) as the ball passes a point A,at one quarter of the maximum height.

1. vA =3 v04

2. vA =

3 v02

correct

3. vA =v02

4. vA =v02

5. vA =

5 v0

22

6. vA =3 v08

7. vA =v04

8. vA =5 v08

9. vA =

3 v0

22

Explanation:

h =hmax4

=v20

8 g, so

v2f = v2

0 + 2 a (y y0)

v2A = v2

0 2 g(v20

8 g

)=

3

4v20

vA =

3

2v0 .

019 10.0 pointsA stone is thrown straight up from the groundwith an initial speed of 44 m/s . At the sameinstant, a stone is dropped from a height ofh meters above ground level. The two stonesstrike the ground simultaneously.Find the height h. The acceleration of

gravity is 9.8 m/s2 .

Correct answer: 395.102 m.

Explanation:

Let : v0 = 44 m/s and

g = 9.8 m/s2 .

Consider the thrown stone

h0 = h0 + v0 T 12g T 2

0 = T

(v0 1

2g T

)

T =2 v

g=

2 (44 m/s)

(9.8 m/s2)= 8.97959 s .

Applying the time to the free fall of the secondstone,

h =1

2g T 2 =

1

2g

(2 v

g

)2= 2

v2

g

= 2(44 m/s)2

9.8 m/s2= 395.102 m .

physics 1443

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