Question

a passanger is standing 'd' meters away from a bus. The bus begins to move with constant acceleration a. to catch the bus , the passanger runs at constant speed V towards the bus. what must be minimum speed of the passanger so that he may catch the bus.

Answer 1

Let us assume they mee at time 't'.
So Bus would have covered a distance of

s= (at^2)/2

So, now the bus is at distance

(s' = d + s) from the passenger.

Now, If Boy has caught the bus, then he would have covered distance-

s' = vt (he runs at constant speed)


Manipulate with these two equations to get your required answer.

Answer 2

Given:

A passenger is standing 'd' meters away from a bus. The bus begins to move with constant acceleration a. to catch the bus, the passenger runs at constant speed V towards the bus.

To find:

What must be the minimum speed of the passenger so that he may catch the bus?

Solution:

Assuming the passenger catches the bus after time "t".

The distance travelled by bus during this time,

x1 = 0 + 1/2at^2  ......(1)

The distance travelled by passenger,

x2 = ut ......(2)

The passenger will catch the bus only when the total distance travelled by the passenger equals distance travelled by bus and the initial distance of separation between the passenger and bus.

d + x1 = x2

using equations (1) and (2), we get,

d + 1/2at^2 = ut

t = [u ±√(u^2 - 2ad)]/2a

assuming the time non-zero and positive, we get,

u^2 ≥ 2ad

u ≥ √2ad

∴ The  minimum speed of the passenger to catch the bus √2ad m/s