Question

Two cars travelling on a straight road cross a kilometre stone A at the same time with velocities 20 m/s and 10 m/s with constant accelerations of 1 m/s^2 and 2 m/s^2 respectively if they cross another kilometre stone B at the same instant the distance between Sand B is

Answer 1

600 Meters

Given:

Velocity of car 1 (v1) = 20 m/sec

Velocity of car 2 (v2) = 10 m/sec

Acceleration of car 1 (a1) = 1 m/sec^2

Acceleration of car 2 (a2) = 2 m/s^2

To Find:

The distance between stone A and Stone B

Solving:

S = Sa = Sb (All three distances are equal)

Formula to find distance:

S = ut + 1/2 at^2

Calculating the distance of Car 1:

Substituting the values known to us in the formula above we get:

S = 20t + 1/2(1)t^2

Calculating the distance of Car 2:

S = 10t + 1/2(2)t^2

S = 10t + t^2

As we know that Sa and Sb are equal Equating them we get:

20t + 1/2 x t^2 = 10t + t^2

t^2/2   = 10t

t = 10 x 2

t = 20 seconds

Putting the value of t in equation 2 we get:

= 10(20) + (20)^2

= 200 + 400

= 600 meters

Therefore, the distance between A and B is 600 meters.