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

Hint: Since the acceleration is a constant show we will use the formula, $ v = u + at $ and from this, we will get the acceleration and since we have to find the distance so by using the formula, $ {v^2} = {u^2} + 2as $ and substituting the values, and solving for the distance, we will get to the answer.

Formula used
Equation of motion given by,
 $ v = u + at $ and
 $ {v^2} = {u^2} + 2as $
Here, $ v $ will be the final velocity,
 $ u $ , will be the initial velocity,
 $ s $ , will be the distance,
 $ a $ , will be the acceleration.

Complete step by step answer:
So in this question, we have the values given as $ u = 30m/s{\text{ , v = 15m/s , t = 3s}} $
So by using the formula $ v = u + at $ we will first find the acceleration. So on substituting the values, we will get to the equation as
 $ \Rightarrow 15 = 30 + a \times 3 $
So on solving the above equation, we will get the value as
 $ \Rightarrow \dfrac{{15 - 30}}{3} = a $
And further solving more, we get
 $ \Rightarrow a = - 5m/{s^2} $
Since in the question they stated that the motorcycle stops so from this we can say that the final velocity will become zero, or mathematically it can be written as
 $ \Rightarrow v = 0 $
So to calculate the displacement we will use the formula $ {v^2} = {u^2} + 2as $ .
So on substituting the values, we get
 $ \Rightarrow {0^2} = {30^2} + 2 \times \left( { - 5} \right)s $
Solving the multiplication part and the power we will get the equation as
 $ \Rightarrow 0 = 900 - 10s $
So taking the constant term at one side, we get
 $ \Rightarrow s = \dfrac{{900}}{{10}}m $
And on solving the division, we get
 $ \Rightarrow s = 90m $
Hence, before stopping the motorcycle it travels $ 90m $ .

Note:
Whenever the object or anybody is set into the motion from the rest then the body when it was resting their initial velocity will become zero. Similarly, if the body stops after applying brakes or anything, as in this case too it happened then in such situations the final velocity becomes zero.