2 cars,one travelling at 54km/hr and other at 90km/hr move towards each other along a narrow road. When they are 150m apart ,both the drivers applythe brakes simultaneously.The motion of each car is retarded at 3m/sec square and the collision is avoided. Find the distance between the cars when they come to rest

Dear Student,

Initial velocity of the car A = u_A= 54km/h = 54x(1000/3600)m/sec = 15m/sec.

Final velocity of car A = v_A = 0

Acceleration of car A = a_A = -3m/sec²

So distance travelled by the car A before coming to rest can be calculated by the following equation-

 v²-u² = 2as

0²-15² = 2x(-3)xs

-225 = -6s

Distance = s = 37.5m

Now Initial velocity of the car B = u_B= 90km/h = 90x(1000/3600)m/sec = 25m/sec.

Final velocity of car B = v_B = 0

Acceleration of car B = a_B = -3m/sec²

So distance travelled by the car B before coming to rest can be calculated by the following equation-

 v²-u² = 2as

0²-25² = 2x(-3)xs

-625 = -6s

Distance = s = 104.16m

Given the Initial distance between the two cars = 150m.

Final distance between the cars when they come to rest = Initial distance – Total distance covered by the two cars.

Final distance between the cars when they come to rest = 150 – (104.16+37.5) = 8.3m

Thus the final distance between the cars when they come to rest will be 8.3m.

Regards.