Question

Question 2
A train is travelling at a speed of $90kmph$. Brakes are applied so as to produce a uniform acceleration of $-0.5m{s}^{-2}$. Find how far the train will go before it is brought to rest.

Answer 1

Step 1: Given that:

The initial speed of the train(u) = $90km{h}^{-1}$ = $90\times \frac{5}{18}m{s}^{-1}=25m{s}^{-1}$

Acceleration in the train(a) = $-0.5m{s}^{-2}$

Final velocity of the train(v) = 0

Step 2: Calculation of the distance the train cover before coming to rest:

Using the third equation of motion;

$v^2-u^2=2as$

Where, v = Final velocity, u= initial velocity, s= distance covered by the body and a= acceleration in the body

Putting the values we get

$0-{\left(25m{s}^{-1}\right)}^2=2\times (-0.5)m{s}^{-2}\times s$

$\Rightarrow -s=-625$

$\Rightarrow s=625m$

Thus,

The train will cover a distance of 625m before coming to rest.