A certain test car can go from rest to 32.0 m/s in 3.88 s.The same car can come to a full stop from that speed in 4.14 s.What is the ratio of the magnitude of the starting acceleration to the stopping acceleration?
A)0.937
B)1.07
C)0.878
D)1.14

Question

Quizplus · Accepted Answer

The starting acceleration (a_start) is calculated using the formula $a = \frac{\Delta v}{\Delta t}$, where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time. For starting, $\Delta v = 32.0 \, \text{m/s}$ and $\Delta t = 3.88 \, \text{s}$, so $a_{start} = \frac{32.0}{3.88}$.The stopping acceleration (a_stop) is calculated similarly, but since the car comes to a stop, $\Delta v = -32.0 \, \text{m/s}$ (the negative sign indicates a decrease in speed) and $\Delta t = 4.14 \, \text{s}$, so $a_{stop} = \frac{-32.0}{4.14}$.The ratio of the magnitudes of these accelerations is $\frac{|a_{start}|}{|a_{stop}|} = \frac{32.0/3.88}{32.0/4.14}$.Calculating both gives $a_{start} \approx 8.247 \, \text{m/s}^2$ and $a_{stop} \approx -7.729 \, \text{m/s}^2$. The ratio is $\frac{8.247}{7.729} \approx 1.07$.

A Certain Test Car Can Go from Rest to 32