Question

A motorcycle rider travelling at 30m/s sees a child run into the streets 190m ahead. If the rider takes 1 second to react before beginning to decelerate at a rate of 3m/s^2, does he stop in time, and if so by what distance is the child safe? (Hint: there a 2 stages of motion)

Answer 1

Answer:

The distance by which the child is safe is  $k = 10 \ m$

Explanation:

From the question we are told that

  The speed of the motorcycle is  $v_n = 30 m/s$

   The distance of the child from the motorcycle is L =  190 m

    The reaction time is  $t_r = 1 \ s$

    deceleration is  $a = -3m/s^2$

Generally the distance covered during the reaction time is  

    $D = v_n * t_r$

=>$D = 30 * 1$

=> $D = 30 \ m$

Generally the distance covered during the deceleration

     $v^2 = u^2 + 2as$

Here v is the final velocity which is  zero  

So

   $0 = 30 ^2 + 2 * (-3)s$

=> $s = 150 \ m$

So the total distance the motorcycle rider will cover before coming to rest is  

     $d = D + s$

=>   $d = 30 + 150$

=>  $d = 180 \ m$

Therefore the amount of distance by which the child id safe is  

    $k = L -d$

=> $k = 190 -180$

=>  $k = 10 \ m$