Question

A car standing 200m behind a bus, which is also at rest. the two start moving at the same instant but with different forward acceleration. the bus has acceleration 2m/s² and the car has acceleration 4m/s². the car will catch up with the bus after a time of?

Answer 1

car can catch the bus within 5 sec ...by using the equation of kinematics ... s= ut + 1/2 at'2 we get ans..

Answer 2

let the car catch up with the bus after the bus covers a distance of xm

for car:

u=0m/s,

a=4m/s^2,

distance=x+200;

from relation,

S=ut+1/2at^2

x+200=2t^2

t^2=(x+200)/2............eq(1)

for bus:

u=0m/s,

a=2m/s^2,

distance=x m

from relation,

S=ut+1/2at^2,

x=t^2.............eq(2)

Eq(1)=eq(2)

x=(x+200)/2

2x=x+200

x=200 m

substituting value of x i eq (1)

t=√200

t=10√2 s