Question

A car and a truck, starting from rest, accelerate at the same rate for time t. The truck continues to accelerate at the same rate until time 2t, while the car coasts. At time 2t, how much further has the truck gone than the car? four times further 1.4 times further twice as far three times as far one third further

Answer 1

Answer:Three times than car

Explanation:

Given

Truck and car starts from rest and accelerates at same rate for t time

Let a be the acceleration

therefore distance traveled by them in t s

$s=ut+\frac{at^2}{2}$

here u=0

$s=\frac{at^2}{2}$

velocity acquired by them in t s

v=u+at

v=at

it is stated that truck continues to accelerate while the car stops

distance traveled in t s

$s_2=vt+\frac{at^2}{2}$

$s_2=at^2+\frac{at^2}{2}$

$s_2=\farc{3at^2}{2}---2$

Divide 1 & 2

$\frac{s_2}{s_1}=\frac{\farc{3at^2}{2}}{\frac{at^2}{2}}$

$\frac{s_2}{s_1}=3$

thus truck has gone  3 times the car has traveled