Question

A boy is x distance behind the car when the car start accelerating from rest with acceleration a with what minimum constant velocity boy start running to catch the car.

Answer 1

Answer:

$\sqrt{2a_0x}$

Explanation:

Let the boy catched the car in time t and his velocity is v

let in time t car moves distance s

then distance covered by boy in time t = x+s

t = (x+s)/v

⇒ x+s = vt

⇒ s = vt - x

acceleration of car = a₀

Thus using the second equation of motion

$s=\frac{1}{2}a_0t^2$

$vt-x=\frac{1}{2}a_0t^2$

$\implies 2vt-2x=a_0t^2$

$\implies a_0t^2-2vt+2x=0$

Which is a quadratic equation in t

But time t will be real

Therefore,

$(2v)^2-4a_0\times2x>0$

$\implies 4v^2-8a_0x>0$

$\implies v^2-2a_0x>0$

$\implies v^2>2a_0x$

$\implies v>\sqrt{2a_0x}$

Hope the answer is helpful.