A train starting from rest attains a velocity of 72 km / h in 5 minute

1.	A train starting from rest attains a velocity of 72 km / h in 5 minutes. Assuming that the acceleration is uniform. find (i) the acceleration and (i) the distance traveled by the train for attaining this velocity -
Answer» GIVEN :- Initial velocity (u) :- 0 m/s Final velocity (v) :- 72 km/hr = 72 × 5/18 = 20 m/s Time (t) :- 5 MINUTES = 5 × 60 seconds To Find :- ACCELERATION (a) Distance travelled (s) Solution :- We know :- $\bf v = u + at$ $: \implies \sf a = \dfrac{v - u}{t}$ $: \implies \sf a = \dfrac{20 - 0}{5 \times 60}$ $: \implies \sf a = \dfrac{20}{300}$ $: \implies \sf a = \dfrac{1}{15}$ Acceleration :- 1/15 m/s^2 We know :- $\bf s = ut + \dfrac{1}{2} a {t}^{2}$ $: \implies \sf s = 0 \times 5 \times 30 + \dfrac{1}{2} \times \dfrac{1}{15} \times 5 \times 60 \times 5 \times 60$ $: \implies \sf s = 5 \times 2 \times 5 \times 60$ $: \implies \sf s = 3000$ Distance travelled :- 3000m = 3 km