Question

A bus is travelling at a speed of 60 km/hour. On seeing a boy 11 m ahead on the road,

the driver of the bus applies brakes and the bus stops at a distance of 10 m. What is its
tardation and how long does it take to come at rest?​

Answer 1

Decelaration is
`13.94m/s^2`

Time is 4.3 seconds

Step-by-step explanation:

Given

`u = 60km/hr` --- Initial Velocity

`s = 10m` --- Distance to stop

`v = 0km/hr` -- Final Velocity

Required

Determine the deceleration (a) and time taken (t)

The deceleration is solved using the following Newton equation of motion

`v^2 = u^2 + 2as`

Convert velocity to m/s

`u = 60km/hr`

`u = (50 * 1000)/(3600) m/s`

`u = 16.7\ m/s`

`v^2 = u^2 + 2as` becomes

`0^2 = 16.7^2 + 2 * a * 10`

`0 = 278.89+ 20 a`

Collect Like Terms

`-20a = 278.89`

Make a the subject

`a = -(278.89)/(20)`

`a = -13.94m/s^2`

The negative sign shows deceleration.

Hence, the decelaration is
`13.94m/s^2`

Solving time (t): Using first law of motion

`v = u + at`

Substitute values for v, u and a

`0 = 60 -13.94 * t`

`0 = 60 -13.94t`

Collect Like Terms

`13.94t = 60`

Solve for t

`t = (60)/(13.94)`

`t = 4.3s`