Question

A car is initially traveling south at a speed of 10.9m/s. The brakes are applied and the car comes to a stop over a distance of 39.2 meters. Find the car's acceleration.

Answer 1

Answer: $-1.51 m/s^{2}$

Explanation:

Since the car is moving in a single direction, we can use the following kinematic equation to find its acceleration $a$:

$V^{2}=V_{o}^{2}+2ad$

Where:

$V=0 m/s$ is the car's final velocity (we are told it comes to a stop)

$V_{o}=10.9 m/s$ is the car's initial velocity

$a$ is the car's acceleration

$d=39.2 m$ is the distance travelled by the car

Isolating $a$:

$a=\frac{V^{2}-V_{o}^{2}}{2d}$

Solving the equation with the given data:

$a=\frac{(0 m/s)^{2}-(10.9 m/s)^{2}}{2(39.2 m)}$

Finally, the acceleration is:

$a=-1.51 m/s^{2}$ Note the negative sign indicates the car was decreasing its velocity before coming to a stop.