Question

A train starting from a railway station and moving with uniform acceleration attains a speed 60 km h–1 in 10 minutes. Find its acceleration.

Answer 1

Given:

Initial velocity of train,u= 0 km/h

(Since, train starts from rest)

Final velocity of train,v= 60 km/h

Time taken by train,t= 10 minutes

To Find:

Acceleration of train

Solution:

We know that,

• According to first equation of motion for constant acceleration,

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

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{t=10\ min=\dfrac{10}{60}\ h=\dfrac{1}{6}\ h}$

$\rule{190}{1}$

Let the acceleration of train be a

So, on applying first equation of motion on train,we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{60=0+a\times\dfrac{1}{6}}$

$\longrightarrow\rm{60=\dfrac{a}{6}}$

$\longrightarrow\rm{\dfrac{a}{6}=60}$

$\longrightarrow\rm\green{a=360\ km/h^{2}}$

Hence, acceleration of train is 360km/h².

Answer 2

Answer: 360 $km/hr^{2}$

Explanation:

Given :-

Initial velocity = $u$ = 0 km/hr (When the train starts, its velocity is 0 km/hr.)

Final velocity = $v$ = 60 km/hr

Time = 10 minutes = 10/60 hr = 1/6 hr

Acceleration = $a$

$a=\dfrac{initial \ velocity - final \ velocity}{time}$ $= \dfrac{v - u}{t}$ $= \dfrac{60 - 0}{\frac{1}{6} }$  

∴ $a$ = 60 × 6 = 360 $km/hr^{2}$