If a velocity of a train which starts from rest is $72 k m / h (20 m / s)$ after $5$ minutes, find out its acceleration and the distance traveled by the train in this time.

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Solution

$u = 0$
$t = 5$ minute
= $5 \times 60 s$
$v = 72 k m / h$
= $20 m / s$

The acceleration is given as:
$a = \frac{v - u}{t}$

= $\frac{20 - 0}{5 \times 60}$

= $\frac{1}{15} m / s^{2}$

$s = u t + \frac{1}{2} a t^{2}$
= $0 \times 5 \times 60 + \frac{1}{2} \times \frac{1}{15} \times 300^{2}$
= $3000 m$

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If a velocity of a train which starts from rest is 72km/h(20m/s) after 5 minutes, find out its acceleration and the distance traveled by the train in this time.

u=0t=5 minute=5×60sv=72km/h=20m/sThe acceleration is given as:a=v−ut=20−05×60=115m/s2s=ut+12at2=0×5×60+12×115×3002=3000m

If a velocity of a train which starts from rest is $72 k m / h (20 m / s)$ after $5$ minutes, find out its acceleration and the distance traveled by the train in this time.