Question

a bus starts from rest and attains a speed of 36km/h^-1 in 10 minutes while moving with uniform acceleration Caculate the acceleration of the bus
( Intial velocity W=0
final velocity V=36km h^-1=36×5÷18ms^-1
time= t=10 minutes
=sec
=a=?
a=v-u÷t​

Answer 1

Provided that:

• Initial velocity = 0 m/s
• Final velocity = 36 km/h
• Time taken = 10 minutes

To calculate:

• The acceleration

Solution:

• The acceleration = 0.016 m/s²

Using concepts:

• Acceleration formula
• Formula to convert kmph-mps
• Formula to convert min-sec

Using formula:

• ${\small{\underline{\boxed{\sf{a \: = \dfrac{v-u}{t}}}}}}$

Where, a denotes acceleration, v denotes final velocity, u denotes initial and t denotes time taken

• ${\small{\underline{\boxed{\sf{1 \: minute \: = 60 \: seconds}}}}}$

Required solution:

~ Firstly let us convert kmph-mps

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let us convert minutes into seconds by using suitable formula!

$:\implies \sf 1 \: minute \: = 60 \: seconds \\ \\ :\implies \sf 10 \: minutes \: = 60 \times 10 \\ \\ :\implies \sf 600 \: seconds \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let's calculate acceleration!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{10-0}{600} \\ \\ :\implies \sf a \: = \dfrac{10}{600} \\ \\ :\implies \sf a \: = \dfrac{1}{60} \: ms^{-2} \\ \\ :\implies \sf a \: = 0.016 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0.016 \: ms^{-2}$