A train starts from rest and attains a velocity of 40 m/s in 10 s with uniform acceleration. What distance does it cover in this time?
200 m
For uniform acceleration, the distance covered equals the average velocity multiplied by the time: \(s = \frac{u + v}{2} \times t\), where u is the initial velocity and v the final velocity.
Here the train starts from rest, so \(u = 0\), the final velocity \(v = 40\) m/s and the time \(t = 10\) s. The average velocity is \(\frac{0 + 40}{2} = 20\) m/s.
So the distance covered is \(s = 20 \times 10 = 200\) m, the train moving with an average speed of 20 m/s throughout the 10 s interval.
Check: the acceleration is \(a = \frac{v - u}{t} = \frac{40}{10} = 4\) m/s2, and \(s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times 10^2 = 200\) m, which matches the earlier result.
Hence, 200 m.
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