# [Physics] Relative motion problem

homework-and-exercisesnewtonian-mechanicsrelative-motion

A man travelling in a car with a maximum constant speed of 20m/s watches the friend start off at a distance 100m ahead on a motorcycle with constant acceleration a. The maximum value of a for which the man in the car can reach his friend is :

(a) 2 m/s2 (b) 4m/s2 (c) 1 m/s2 (d) None of these.

I was trying to solve this question and all I could do was this. Please give me a hint on how to solve this.

If the motorcycle starts at $x=0$ at $t=0s$, then the position of the motorcycle is given by

$x = 0.5·a·t^2$, dependent on time $t$.

The car starts at $x = -100m$ at $t=0s$ and is on time $t$ at

$x = -100m+vt$

for the velocity $v = 20m/s$. Now equate these equations to get

$-100m + vt = x = 0.5at^2$.

You can solve this quadratic equation, but maybe it can occur that you get negative numbers in square root, something which would be unphysical. The value of the discriminant (number under square root) must be positive; therefore you must search, when the discriminant is Zero (on this value of $v$ the Transition to unphysical solution, the negative discriminant, takes place).

That condition allows you to solve by $v$.