Question

A passenger standing 20m behind the bus the bus begins to move with. Constant acceleration

Answer 1

Considering the question to be,

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. what must be the minimum speed of the passenger so that he may catch the bus.

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

From given, we have, d = 20 m

The minimum speed of the passenger so that he may catch the bus is √(2 × a × 20) = √40a

∴ The minimum speed of the passenger so that he may catch the bus is 6.32a m/s