Question

A train travelling at an average speed of 60 km/h comes to a complete stop in 20 minutes' time. What is its deceleration?

Answer 1

GIVEN :

• Initial speed of train, u = 60 km/h
• final speed of train, v = 0  [ since it comes to rest ]
• time taken by train to stop, t = 20 min = 20/60  hrs = 1/3 hrs

TO FIND :

• Deceleration of the Train

FORMULA REQUIRED :

• First equation of motion

       v = u + a t

[ where v is final velocity, u is initial velocity, a is acceleration and t is time taken ]

SOLUTION :

• Deceleration is known as negative acceleration.

so, Let acceleration of train be a

Using first equation of motion

→ v = u + a t

→ 0 = 60 + a ( 1 / 3 )

→ - 60 = a / 3

→ a = - 60 × 3

→ a = -180  km/h²

therefore,

Acceleration of Train is  -180 km/h², hence

• Deceleration of Train is 180 km/h²

THREE EQUATIONS OF MOTION :

• First equation of motion

     v = u + a t

• Second equation of motion

     s = u t + 1/2 a t²

• Third equation of motion

     2 a s = v² - u²

[ where v is final velocity, u is initial velocity, a is acceleration, t is time taken and s is distance covered by body]

Answer 2

Given ,

Initial velocity (u) = 60 Km/hr

Final velocity (v) = 0 Km/hr

Time (t) = 20 min or 0.3 hr

We know that ,

The negative acceleration is called deceleration

And

$\boxed{ \tt{Acceleration \: ( a) = \frac{v - u}{t} }}$

Thus ,

a = (0 - 60)/0.3

a = -60/0.3

a = -200 Km/hr²

∴The deceleration of object is 200 km/hr²