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

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

Step-by-step 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`