A metro train starts from a station and under uniform acceleration, it acquires a velocity of $72 k m / h$ in $10$ seconds. Calculate its acceleration and the distance covered during this time.

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Solution

Step 1 - Given
Initial velocity $(u) = 0 m / s .$
Final velocity $(v) = 72 k m / h = 72 \times \frac{1000 (m)}{3600 (s)} = 20 m / s . (1 k m = 1000 m a n d 1 h r = 3600 s e c)$
Time taken = $10$ seconds.
Step 2 - Calculation of acceleration
Acceleration is defined as the rate of change in velocity.
Formula for acceleration is given by. $a c c e l e r a t i o n = \frac{f i n a l v e l o c i t y (v) - i n i t i a l v e l o c i t y (u)}{t i m e t a k e n (t)}$
Substituting the values in the given relation.
$a = \frac{20 - 0}{10} = 2 m / s^{2} .$
Step 3 - Calculation of distance:
For uniformly accelerated motion, the position-velocity relation is given by: $v^{2} = u^{2} + 2 a s .$
Substituting the values in the given relation.
${(20)}^{2} = {(0)}^{2} + 2 \times 2 \times s$
$400 = 4 \times s$
$s = \frac{400}{4} = 100 m .$
Hence, the acceleration is $2 m / s^{2}$ and the distance covered is $100 m$ .

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A metro train starts from a station and under uniform acceleration, it acquires a velocity of 72km/h in 10 seconds. Calculate its acceleration and the distance covered during this time.

A metro train starts from a station and under uniform acceleration, it acquires a velocity of $72 k m / h$ in $10$ seconds. Calculate its acceleration and the distance covered during this time.