Question

A bus starting from a station attains a velocity of  36 km/h in 10 seconds. Assuming that the acceleration  is uniform, find

(i) The acceleration.

(ii) Distance travelled by the bus for attaining this  velocity.​

Answer 1

Answer:

Acceleration = 1 m/s²

Distance = 50 m

Explanation:

As per the provided information in the given question, we've :

• Initial velocity (u) = 0 m/s (As it was at rest first)
• Final velocity (v) = 36 km/h
• Time taken (t) = 10 s

In order to calculate acceleration and distance travelled by the bus for attaining this  velocity, firstly we need to convert final velocity in its standard form, that is m/s.

$\longmapsto \rm {1 \; kmh^{-1} = \dfrac{5}{18} \; ms^{-1} }$

$\longmapsto \rm {36 \; kmh^{-1} = \Bigg ( \dfrac{5}{18} \times 36 \Bigg ) \; ms^{-1} }$

$\longmapsto \rm {36 \; kmh^{-1} = ( 5\times 2 ) \; ms^{-1} }$

$\longmapsto \rm {36 \; kmh^{-1} = 10 \; ms^{-1} }$

$\longmapsto \bf {Final \; velocity = 10 \; ms^{-1} }$

$\rule{200}2$

Calculating acceleration of the bus (a) :

By using the first equation of motion :

$\longmapsto \bf {v = u + at}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto \rm {10 = 0 + 10a}$

$\longmapsto \rm{10 -0= 10a}$

$\longmapsto \rm {10 = 1 0a}$

$\longmapsto \rm {\cancel{\dfrac{10}{10}} = a}$

$\longmapsto \bf {1 \; m/s^2 = a}$

∴ Acceleration of the bus is 1 m/s².

$\rule{200}2$

Calculating distance travelled by the bus for attaining this  velocity (s) :

By using the third equation of motion,

$\longmapsto \bf {v^2 -u^2 = 2as}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longmapsto \rm {(10)^2 -(0)^2 = 2\times 1 \times s}$

$\longmapsto \rm {100 -0 = 2s}$

$\longmapsto \rm {100 = 2s}$

$\longmapsto \rm {\cancel{\dfrac{100}{2}} = s}$

$\longmapsto \bf {50 \; m= s}$

∴ Distance travelled by the bus for attaining this  velocity is 50 m.