Problem 2D Ch. 2–7

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Holt Physics

Problem 2DVELOCITY AND DISPLACEMENT WITH UNIFORM ACCELERATION

P R O B L E MA barge moving with a speed of 1.00 m/s increases speed uniformly, sothat in 30.0 s it has traveled 60.2 m. What is the magnitude of the barge’sacceleration?

S O L U T I O N

Given: vi = 1.00 m/s

∆t = 30.0 s

∆x = 60.2 m

Unknown: a = ?

Use the equation for displaement with constant uniform acceleration.

∆x = vi∆t + 12

a∆t2

Rearrange the equation to solve for a.

12

a∆t2 = ∆x − vi∆t

a = 2(∆x

∆−t2

vi∆t)

a =

a =

a = 9

(

.

2

0

)

0

(3

×0

1

.2

02m

s

)2

a = 6.71 × 10−2 m/s2

(2)(60.2 m − 30.0 m)

9.00 × 102 s2

(2)[60.2 m − (1.00 m/s)(30.0 s)]

(30.0 s)2

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ADDITIONAL PRACTICE

1. The flight speed of a small bottle rocket can vary greatly, depending on

how well its powder burns. Suppose a rocket is launched from rest so that

it travels 12.4 m upward in 2.0 s. What is the rocket’s net acceleration?

2. The shark can accelerate to a speed of 32.0 km/h in a few seconds. As-

sume that it takes a shark 1.5 s to accelerate uniformly from 2.8 km/h to

32.0 km/h. What is the magnitude of the shark’s acceleration?

3. In order for the Wright brothers’ 1903 flyer to reach launch speed, it had

to be accelerated uniformly along a track that was 18.3 m long. A system

of pulleys and falling weights provided the acceleration. If the flyer was

initially at rest and it took 2.74 s for the flyer to travel the length of the

track, what was the magnitude of its acceleration?

Holt Physics Problem BankCh. 2–8

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4. A certain roller coaster increases the speed of its cars as it raises them to

the top of the incline. Suppose the cars move at 2.3 m/s at the base of

the incline and are moving at 46.7 m/s at the top of the incline. What is

the magnitude of the net acceleration if it is uniform acceleration and

takes place in 7.0 s?

5. A ship with an initial speed of 6.23 m/s approaches a dock that is 255 m

away. If the ship accelerates uniformly and comes to rest in 82 s, what is

its acceleration?

6. Although tigers are not the fastest of predators, they can still reach and

briefly maintain a speed of 55 km/h. Assume that a tiger takes 4.1 s to

reach this speed from an initial speed of 11 km/h. What is the magni-

tude of the tiger’s acceleration, assuming it accelerates uniformly?

7. Assume that a catcher in a professional baseball game catches a ball that

has been pitched with an initial velocity of 42.0 m/s to the southeast. If

the catcher uniformly brings the ball to rest in 0.0090 s through a dis-

tance of 0.020 m to the southeast, what is the ball’s acceleration?

8. A crate is carried by a conveyor belt to a loading dock. The belt speed

uniformly increases slightly, so that for 28.0 s the crate accelerates by

0.035 m/s2. If the crate’s initial speed is 0.76 m/s, what is its final speed?

9. A plane starting at rest at the south end of a runway undergoes a uni-

form acceleration of 1.60 m/s2 to the north. At takeoff, the plane’s ve-

locity is 72.0 m/s to the north.

a. What is the time required for takeoff?

b. How far does the plane travel along the runway?

10. A cross-country skier with an initial forward velocity of +4.42 m/s ac-

celerates uniformly at −0.75 m/s2.

a. How long does it take the skier to come to a stop?

b. What is the skier’s displacement in this time interval?

Section Five—Problem Bank V Ch. 2–5

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.Givens Solutions

2. ∆t = 1.5 s

vi = 2.8 km/h

vf = 32.0 km/ha = =

a = = 5.4 m/s2

(29.2 km/h)�36

1

0

h

0 s��11

0

k

3

m

m�

1.5 s

(32.0 km/h − 2.8 km/h)�36

1

0

h

0 s��10

k

3

m

m�

1.5 s

vf − vi∆t

3. ∆x = 18.3 m

∆t = 2.74 s

vi = 0 m/s

Because vi = 0 m/s, a = �2

∆∆t

x2� =

(2

(

)

2

(

.

1

7

8

4

.3

s)

m2

) = 4.88 m/s2

4. vi = 2.3 m/s

vf = 46.7 m/s

∆t = 7.0 s

a = = 46.7 m/

7

s

.0

−s

2.3 m/s =

44

7

.4

.0

m

s

/s = 6.3 m/s2vf − vi

∆t

5. vi = 6.23 m/s

∆x = 255 m

∆t = 82 s

a = 2(∆x

∆−t2vi ∆t) =

a = = �(

6

2

.

)

7

(−×2

1

5

0

53m

s2)

� = −7.6 × 10−2 m/s2(2)(255 m − 510 m)

6.7 × 103 s2

(2)[255 m − (6.23 m/s)(82 s)]

(82 s)2

6. vi = 11 km/h

vf = 55 km/h

∆ = 4.1 sa = =

a = = 3.0 m/s2

(44 km/h)�36

1

0

h

0 s��11

0

k

3

m

m�

4.1 s

(55 km/h − 11 km/h)�36

1

0

h

0 s��11

0

k

3

m

m�

4.1 s

vf − vi∆t

7. vi = 42.0 m/s southeast

∆t = 0.0090 s

∆x = 0.020 m/s southeast

a = 2(∆x

∆−t2vi ∆t) =

a = =

a = −8.9 × 103 m/s2, or 8.9 × 103 m/s2 northwest

(2)(−0.36 m)8.1 × 10−5 s2

(2)(0.020 m/s − 0.38 m)

8.1 × 10−5 s2

(2)[0.020 m − (42.0 m/s)(0.0090 s)]

(0.0090 s)2

8. ∆t = 28 s

a = 0.035 m/s2

vi = 0.76 m/s

vf = a∆t + vi = (0.035 m/s2)(28.0 s) + 0.76 m/s = 0.98 m/s + 0.76 m/s = 1.74 m/s

Additional Practice 2D

1. ∆x = 12.4 m upward

∆t = 2.0 s

vi = 0 m/s

Because vi = 0 m/s, a = 2

∆∆t2x

= (2)

(

(

2

1

.0

2.

s

4

)2m)

= 6.2 m/s2 upward

Givens Solutions

Holt Physics Solution ManualV Ch. 2–6

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.

4. ∆x = 2.00 × 102 m

vi = 9.78 m/s

vf = 10.22 m/s

a = vf

2

2∆−

x

vi2

= =

a = 4.

8

0

.

0

8

×m

1

2

0

/2s2

m = 2.2 × 10−2 m/s2

104.4 m2/s2 − 95.6 m2/s2

4.00 × 102 m

(10.22 m/s)2 − (9.78 m/s)2

(2)(2.00 × 102 m)

10. vi = +4.42 m/s

vf = 0 m/s

a = −0.75 m/s2

∆t = 5.9 s

a. ∆t = vf −

a

vi = 0 m

−/

0

s

.7

−5

4

m

.42

/s2m/s

= −−0

4

.

.

7

4

5

2

m

m

/

/

s

s2 =

b. ∆x = vi∆t + 1

2a∆t2 = (4.42 m/s)(5.9 s) +

1

2(−0.75 m/s2)(5.9 s)2

∆x = 26 m − 13 m = 13 m

5.9 s

1. vi = 1.8 km/h

vf = 24.0 km/h

∆x = 4.0 × 102 ma =

vf2

2∆−

x

vi2

=

a =

a = = 5.5 × 10−2 m/s2(573 km2/h2)�36

1

0

h

0 s�

2

�1

1

0

k

3

m

m�

2

8.0 × 102 m

(576 km2/h2 − 3.2 km2/h2)�36

1

0

h

0 s�

2

�1

1

0

k

3

m

m�

2

8.0 × 102 m

[(24.0 km/h)2 − (1.8 km/h)2]�36

1

0

h

0 s�

2

�1

1

0

k

3

m

m�

2

(2)(4.0 × 102 m)

2. vf = 0 m/s

vf = 8.57 m/s

∆x = 19.53 m

a = vf

2

2

−∆x

vi2

= = 7

3

3

9

.4

.0

m

6

2

m

/s2

= 1.88 m/s2(8.57 m/s)2 − (0 m/s)2

(2)(19.53 m)

9. vi = 0 m/s

vf = 72.0 m/s north

a = 1.60 m/s2 north

∆t = 45.0 s

a. ∆t = vf −

a

vi = 72.0

1.

m

60

/s

m

−/

0

s2m/s

=

b. ∆x = vi∆t + 1

2a∆t2 = (0 m/s)(45.0 s) +

1

2 (1.60 m/s2)(45.0 s)2 = 0 m + 1620 m

∆x = 1.62 km

45.0 s

Additional Practice 2E

3. vi = 7.0 km/h

vf = 34.5 km/h

∆x = 95 ma =

vf2

2

−∆x

vi2

=

a =

a = = 0.46 m/s2

(1140 km2/h2)�36

1

0

h

0 s�

2

�1

1

0

k

3

m

m�

2

190 m

(1190 km2/h2 − 49 km2/h2)�36

1

0

h

0 s�

2

�1

1

0

k

3

m

m�

2

190 m

[(34.5 km/h)2 − (7.0 km/h)2]�36

1

0

h

0 s�

2

�1

1

0

k

3

m

m�

2

(2)(95 m)