NEET2016PhysicsMotion in One DimensionActual
A man is at a distance of 6 ~m from a bus. The bus begins to move with a constant acceleration of 3 ~m ~s ⁻² . In order to catch the bus, the minimum speed with which the man should run towards the bus is
Options
- A2 ~m ~s ⁻¹
- B4 ~m ~s ⁻¹
- C6 ~m ~s ⁻¹
- D8 ~m ~s ⁻¹
Correct answer
C. 6 ~m ~s ⁻¹
Step-by-step solution
If the man did not run, the bus would be at a distance s₁ at time t given by s₁=6+ 1 2 a t^2=6+ 1 2 3 t^2=6+ 3 2 t^2 If v is the speed of man, he would cover a distance s₂=v t in time t . To catch the bus, s₁=s₂6+ 3 2 t^2=v t or t^2- 2 v 3 t+4=0 which gives t= 2 v 6 1 2 [ 4 v^2 9 -16 ]^ 1 / 2 Now, t will be real if ( 4 v^2 9 -16 ) is positive or zero. Minimum v corresponds to 4 v^2 9 -16=0 which gives v=6 ~m ~s ⁻¹ .