Question

A train running at 30 metre per second is slowed uniformly to stop in 44 second find the acceleration and the stopping distance.

Answer 1

Given : -

•Initial velocity of train is = 30m/s

•Time taken to stop the train is = 44 second

•Final velocity of the train is = 0

To Find : -

•Find the acceleration and the stopping distance.

Solution : -

Given,

• u = 30m/s
• t = 44 sec
• v = 0

we have,

$\: \: \sf \large \implies a = \frac{v - u}{t}$

Now put the values

$\: \sf \implies \: a = \frac{0 - 30}{44} \\ \\ \: \sf \implies \: a = - 0.68 \frac{m}{ {s}^{2} }$

Hence,acceleration of the train is -0.68m/s^2.

Now,we have to find the question that stopping distance of the train.

$\: \sf \large \implies \: s = ut + \frac{1}{2} a {t}^{2}$

Now put the values

$\: \sf \implies \: s = (30)(44) + \frac{1}{2} ( - 0.68) {(44)}^{2} \\ \\ \: \sf \implies \: s = 662m$

Thus,distance covered to stop the train is 662m.

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