A bus starts from rest and attains a speed of $36 k m / h$ in $1 m i n u t e$ while moving with uniform acceleration along a straight road. Calculate the acceleration of the bus.

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Solution

Step 1 - Given:
Initial velocity $(u)$ = $0 m / s$ .
Final velocity $(v)$ = $36 k m / h$ = $36 \times \frac{1000 (m)}{3600 (s)} = 10 m / s .$ $(1 k m = 1000 m a n d 1 h r = 3600 s e c .)$
Time taken $(t)$ = $1 m i n u t e$ = $60 s e c o n d s$ .
Step 2 - Calculation of acceleration:
Acceleration is defined as the rate of change of velocity.
The formula for acceleration is given by,
$a c c e l e r a t i o n = \frac{f i n a l v e l o c i t y (v) - i n i t i a l v e l o c i t y (u)}{t i m e t a k e n (t)}$
$a = \frac{10 - 0}{60} = \frac{10}{60} = 0.166 m / s^{2} .$
Hence, the acceleration of the bus is $0.166 m / s^{2} .$

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