Q105P - Fundamentals Of Physics

Question

Question: A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.0 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops?

Step-by-Step Solution

Answer

The distance traveled by the motorcycle during deceleration is 90 m.

Given Data

Initial speed is, $v_{0=30\text{m/s}}$

Final speed after 3.0 s is, $v=15 \text{m/s}$ 

Understanding of the kinematic equations.

 

Kinematic equations describe the motion of an object with constant acceleration. Using these equations the deceleration of the motorcycle on applying breaks can be computed. This deceleration can be used to find the distance traveled by the motorcycle.  

The expression for the kinematic equations of motion are given as follows: 

     ${\text{v=v}}_0\text{+at}$                                                                         … (i)

 ${\text{v}}^{2 }{{\text{= v}}_0}^2\text{+2ax}$

                                                                                        … (ii)

 

Here,${\text{v}}_0$  is the initial velocity,  $t$ is the time,  $a$is the acceleration and $x$ is the displacement.

 

Determination of the deceleration of motorcycle

Using equation (i), the acceleration can be calculated as, 

 $$\begin{gathered} \text{a=}\frac{{\text{v-v}}_0}{t} \\ \text{ =}\frac{15\text{m/s-30m//s}}{3.0\text{s}} \\ {\text{= -5m/s}}^2 \end{gathered}$$

 

Since acceleration is negative, it is called deceleration

Determination of the distance travelled

Using the equation (ii) the distance traveled can be calculated as,

 $$\begin{gathered} x=\frac{v^2-{v_0}^2}{2a} \\ = \frac{{\left(0\text{m/s}\right)}^2-{\left(30\text{m/s}\right)}^2}{2\times \left(-5{\text{m/s}}^2\right)} \\ = 90\text{m} \end{gathered}$$

 Thus the distance travelled during deceleration is $90\text{m}$ .