Question

A train is travelling at a speed of 90 kilometre per hour. breakes are applied so as to produce a uniform acceleration of -0.5 metre per second square. Find how far the train will go before it is brought to rest.​

Answer 1

$\large{\tt{Answer\ :}}$

$\underline{\text{Given}}$;

u (Initial Velocity) = 90 km/h-¹ = 90 × 5/18 = 25 m/s-¹

v (Final Velocity)= 0

a (Acceleration) = -0.5 m/s-²

$\underline{\text{To find}}$: "S"

Using the 2nd equation of Motion,

=> v² - u² = 2aS

=> (0)² - (25)² = 2 (-0.5) × S

=> $\dfrac{-625}{2× -0.5}$

= 625m

So, "S" = 625 m

So, the train will go $\bold{625\ m}$ before it is brought to rest.

Answer 2

Solution:-

Given:-

u = 90 km/hr.

v = 0 m/s²

a = - 0.5 m/s²

To Find :-

How far the train will go before it is brought to rest.

Find:-

Let "d" be the Distance, that train will go before it is brought to rest.

By the Formula,

v²-u²=2as

=) 0² - 25² = 2( - 0.5 × d)

=) 625 = - 1.0 × d

=) d = 625m.

Hence,

The Train will go upto 625m before it is brought to rest.