Question

You are driving through town at 12.0 m/s when suddenly a ball rolls out in front of you. You apply the brakes and begin slowing at a rate of 3.5 m/s2. How far did you travel before stopping? What variable represents 3.5 m/s2?

Answer 1

Answer:

20.6 m, acceleration

Explanation:

Since this is a uniformly accelerated motion, we can solve the problem with the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity of the driver (since he comes to a stop)

u = 12.0 m/s is the initial velocity

$a=-3.5 m/s^2$ is the acceleration, which is negative since the car is slowing down (so its direction is opposite to the motion)

s is the distance covered by the car while stopping

Solving the equation for s, we find the distance covered:

$s=\frac{v^2-u^2}{2a}=\frac{12.0^2-0}{2(-3.5)}=20.6 m$

And as we said previously, $3.5 m/s^2$ is the magnitude of the acceleration, which represents the rate of change of the velocity.