A student is at a distance 16 ~m from a bus when the bus begins to move with a constant acceleration of 9 ~m ~s ⁻² . The minimum velocity with which the student should run towards the bus so as the catch it is 2 ~ms ⁻¹ . The value of is

Step-by-step solution

Let v be the minimum velocity of student so,that he could catch the bus If student catch the bus in time t , then distance travelled by student in time t=16+ distance travelled by bus in time t . aligned & v t=16+ (u t+ 1 2 a t^2 ) & v t=16+0 t+ 1 2 9 t^2 & v t=16+ 9 2 t^2 9 t^2-2 v t+32=0 aligned The above equation must have real roots. i.e its discriminant 0 i.e (2 v)^2-4 9 32 0 aligned & 4 v^2-4 288 0 v^2-288 0 & v^2 288 aligned v 12 2 ~m / s Minimum velocity of student to catch the bus =12 2 ~m / s = 2 ~m / s (