Question

A train starts from a station and moves with a constant acceleration for 2 minutes. If it covers a distance of 400m in this period, find the acceleration?​

Answer 1

Answer:-

The acceleration =$\bf{0.0556 \: m/s^2}$

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• Given:-

Initial velocity [u] = $\bf{0}$

Time taken [t] = $\bf{2 \: min.}\implies \bf{120 \: sec.}$

Distance covered = $\bf{400 \: m}$

• To Find:-

Acceleration = $\bf{?}$

• Solution:-

Using 2nd Equation of Motion:-

$\bf\boxed{s = ut + \dfrac{1}{2} a t ^2}$

$\implies 400 = 0 \times 120 + \dfrac{1}{2} a \times (120)^2$

$\implies 400 = 0 + \dfrac{1}{2} a \times 14400$

$\implies 400 = a \times 7200$

$\implies a = \dfrac{400}{7200}$

$\implies a = \cancel{\dfrac{400}{7200}}$

$\implies a = \dfrac{1}{18}$

$\implies a = \bf{0.05555}$

$\implies\bf{ a = 0.0556} \: m/s^2$

Therefore,

Acceleration is $\bf{0.0556 \: m/s^2}$

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Equations Of Motion:-

• 1st equation of motion:-

$\bf\boxed{v = u + at}$

• 2nd equation of motion:-

$\bf\boxed{s = ut + \dfrac{1}{2} a t ^2}$

• 3rd equation of motion:-

$\bf\boxed{v^2 - u^2 = 2as}$

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