Question

The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.
i. Compute the acceleration.
ii. How much further will the Bus travel before coming to rest, provided the acceleration remain constant?

Answer 1

Given:

The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.

To find:

• Acceleration of the bus ?
• Distance travelled further to come to rest ?

Calculation:

Let acceleration be "a":

According to 3rd equation of kinematics:

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$\implies\: {(7)}^{2} = {(15)}^{2} + 2a(90)$

$\implies\: 49= 225 + 180a$

$\implies\: 180a = - 176$

$\implies\: a = - 0.97 \: m {s}^{ - 2}$

So, acceleration of the car is -0.97 m/s², negative sign denotes deceleration.

Let the distance travelled further be "d":

$\therefore \: {v}^{2} = {u}^{2} + 2ad$

$\implies\: {(0)}^{2} = {(7)}^{2} + 2( - 0.97)d$

$\implies\: 0= 49 - 1.94d$

$\implies\: 1.94d = 49$

$\implies\: d = \dfrac{49}{1.94}$

$\implies\: d = 25.25 \: m$

So, further distance travelled before coming to rest is 25.25 metres.