Answer to Question #125267 in Physics for Sejal

Question

A bike is traveling to the left with a speed of 27 (m)/(s). when the rider slams on the brakes. The bike skids for 41.5m with constant acceleration before it comes to a stop. What was the acceleration of the bike as it came to a stop?

Answer 1

The kinematic law of motion under the constant acceleration is given be the following equation:

$$d = v_0t + \dfrac{at^2}{2}$$

where $d = 41.5m$ is the traveled distance, $v_0 = 27 m/s$ is the initial velocity, $a$ is the constant acceleration and $t$ is the time of motion.

By definition, the acceleration is given by:

$$a = \dfrac{0-v_0 }{t} = -\dfrac{v_0 }{t}$$

where 0 stands for the final velocity (bike stops completely after time $t$ passes).

Substituting this expression to the law of motion, get:

$$d = v_0t - \dfrac{v_0t}{2} = \dfrac{v_0t}{2}$$

Expressing $t$ and substituting it to the expression for the acceleration, obtain:

$$t = \dfrac{2d}{v_0}\\ a = -\dfrac{v_0^2}{2d}$$

Calculating the numerical value, get:

$$a = -\dfrac{27^2}{2\cdot 41.5}\approx -8.78 m/s^2$$

Answer. The acceleration is -8.78 m/s^2.