A train starting from rest attains a velocity of $72 km / h$ in $5 m i n u t e s$ . Assuming that the acceleration is uniform, find acceleration of train.

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Solution

Assmuptions:
$1 m i n u t e = 60 s e c$
$1 k m / h = \frac{5}{18} m / s$
Step 1: Given Data:
Initial velocity, $= u = 0 m / s$
Final velocity of train, $v = 72 k m / h = \frac{5}{18} \times 72 = 20 m / s$
Time taken by train to cover certain distance, $t = 5 m i n u t e s = 5 \times 60 = 300 s e c$
Step 2: Acceleration of train:
As we know from first equation of motion,
$v = u + a t 20 m / s = 0 + a (300 s e c) \Rightarrow a = \frac{20}{300} = \frac{1}{15} m / s^{2}$
Hence, Acceleration of train, $= \frac{1}{15} m / s^{2}$

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