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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 aavg = -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.                                                                aavg = Dv/Dt  becomes  Dt = Dv/ aavg . So Dt = (-9.0 m/s)/( -4.1 m/s˛) = 2.2 s

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  aavg = 2.5 m/s˛ , so plug in those values into the formula Dt = Dv/ aavg .                                                                                                                                                      Dt = (5m/s)/(2.5 m/s˛) = 2 s

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/ aavg  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. aavg = -0.50 m/s˛, so plug in the values.                                                                                                   Dt = (-13.5m/s)/(-0.50 m/s˛) = 27 s

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 aavg = 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                 aavg = (-5.3m/s)/(1500s) = -3.5 * 10 ^ -3 m/s˛