Practice 2B: | 1
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**Average acceleration- by
Jacobson˛ , 2005**

**1. As the shuttle bus comes to a sudden stop to avoid
hitting a dog, it accelerates uniformly at -4.1 m/s˛ as it slows from 9.0 m/s
to 0 m/s. Find the time interval of acceleration for the bus. **

Since we know a_{avg }= -4.1 m/s˛ and Dv = -9.0 m/s (v
final – v initial), we can rearrange the average acceleration formula to solve
for t. a_{avg}
= Dv/Dt becomes
Dt = Dv/ a_{avg }. So Dt
= (-9.0 m/s)/( -4.1 m/s˛) = **2.2 s**

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**2. A car traveling at 7.0 m/s accelerates uniformly at 2.5
m/s˛ to reach a speed of 12 m/s. How long does it take for this acceleration to
occur? **

Dv
= v final – v initial = 12 – 7 = 5 m/s
and a_{avg }= 2.5 m/s˛ , so plug in those values
into the formula Dt = Dv/ a_{avg . }Dt = (5m/s)/(2.5 m/s˛) =
**2 s**

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**3. With an average acceleration of -0.50 m/s˛, how long
will it take a cyclist to bring a bicycle with an initial speed of 13.5 m/s to
a complete stop? **

Again, use the formula Dt = Dv/ a_{avg } to find the answer. We know that v final = 0
m/s and that v initial is 13.5 m/s, so Dv
= v final – v initial = 0 – 13.5 = -13.5m/s. a_{avg }= -0.50 m/s˛, so plug in the values.
Dt = (-13.5m/s)/(-0.50 m/s˛) = **27 s**

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**4. Turner’s treadmill runs with a velocity of -1.2 m/s and
speeds up at regular intervals during the half-hour workout. After 25 min, the
treadmill has a velocity of -6.5 m/s. What is the average acceleration of the
treadmill during this period?**

Using the average acceleration formula a_{avg} = Dv/Dt , plug in the values that you know. Dv = -6.5m/s - -1.2m/s = -5.3m/s and Dt = (25min)(60sec/min) = 1500 s , so a_{avg} =
(-5.3m/s)/(1500s) = **-3.5 *****
10 ^ -3 m/s˛**

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**5. Suppose a treadmill has an average acceleration of 4.7 *****
10 ^ -3 m/s˛. a. How much does its speed change after 5.0
min? b. If the
treadmill’s initial speed is 1.7 m/s, what will its final speed be?**

a. First rearrange the average acceleration formula to solve for
velocity. a_{avg} = Dv/Dt becomes
Dv = Dt a_{avg }. Dt
= (5.0min)(60s/min) = 300s , so Dv
= (300s)(4.7 * 10 ^ -3 m/s˛) = **1.4 m/s**

b. Since Dv = v_{f }- v_{i
}, you can solve for the final velocity by adding the initial velocity and
the change in velocity you found in part a.
v_{f }= 1.7 m/s + 1.4 m/s = **3.1
m/s**

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