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) 1.07
B) 0.937
C) 1.14
D) 0.878

Question

Quizplus · Accepted Answer

The starting acceleration (a1) can be found 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, $\Delta v = 32.0 \, m/s$ and $\Delta t = 3.88 \, s$, so $a1 = \frac{32.0}{3.88}$. The stopping acceleration (a2) is similarly found with $\Delta v = -32.0 \, m/s$ (since it's decelerating) and $\Delta t = 4.14 \, s$, so $a2 = \frac{-32.0}{4.14}$. The ratio of the magnitudes of these accelerations is $|\frac{a1}{a2}|$, which simplifies to $|\frac{32.0/3.88}{-32.0/4.14}|$, giving a ratio of approximately 1.07.

A Certain Test Car Can Go from Rest to 32