Question

how long would it take a 3.4 kg bike with 79 kg rider traveling at 18.6 m/s to stop if 867 N of force is applied to the brakes?

Answer 1

Answer: 1.76 s

Explanation:

We have the following data:

$m=3.4 kg+79 kg=82.4 kg$ is the total mass of the bike and the rider

$V_{o}=18.6 m/s$ is the initial velocity

$F=867 N$ is the force applied to the brakes

Firstly, we will find the acceleration $a$ with the following equation:

$F=m.a$ (1)

Isolating $a$:

$a=\frac{F}{m}$ (2)

$a=\frac{867 N}{82.4 kg}$ (3)

$a=10.52 m/s^{2}$ (4) This is the magnitude of the acceleration, however, since the final velocity is 0 m/s, this means the direction is negative

Hence:

$a=-10.52 m/s^{2}$ (5)

On the other hand, with the following equation we can find the time $t$:

$V=V_{o}+at$ (6)

Where:

$V=0 m/s$ is the final velocity (the bike stops)

Isolating $t$:

$t=-\frac{V_{o}}{a}$ (7)

$t=-\frac{18.6 m/s}{-10.52 m/s^{2}}$ (8)

Finally:

$t=1.76 s$ This iste time it takes to the bike to stop