Question

A train initially at rest, starts moving and attains a velocity of 72km/hr in 4 minutes. Find :(i) acceleration , (ii) distance covered by the train. {Give answer in m/s terms and dont leave answer in fraction. }

Answer 1

Answer:

Explanation:

v=u+at

a=(v-u)/t -----------(1)

v=72 km/h

=20m/s

u=0

t=5*60 sec =300sec

by putting values of v,u,t in (1)

a=1/15 m/s2

s=ut +(1/2)at2

=0+(1/2)(1/15)(300)(300)

=3km

Answer 2

$\boxed{\bf{ \blue{\bigstar Find \longrightarrow }}}$

(i) acceleration

(ii) distance covered by the train.

$\boxed{\bf{ \red{\bigstar Given \longrightarrow }}}$

Initial velocity = 0m/s

Final velocity = 72km/h

Time = 4 sec

$\boxed{\bf{ \pink{\bigstar Formula \ used \longrightarrow }}}$

$\sf{a= \dfrac{v-u}{t} }$

$\sf{s= ut + \dfrac{1}{2} at^2}$

where a= acceleration

v= final velocity

u= initial velocity

t= time

s= distance

$\boxed{\bf{ \green{\bigstar SoLution \longrightarrow }}}$

Firstly change speed km/h to m/s and time min to sec,

$\sf{Final \ velocity = 72 km/h = \dfrac{5}{18} \times 72 \ m/s = 20m/s}$

$\sf{Time = 4 min. = 4 \times 60 = 240sec}$

Now, for finding the acceleration[a] , use this formula,

$\sf{a= \dfrac{v-u}{t} = \dfrac{20-0}{240} = 0.083 m/s^2}$

For distance covered use second eq' of motion,

$\sf{s= ut + \dfrac{1}{2} at^2}$

$\sf{s=0 \times 240 + \dfrac{1}{2} 0.083 \times 240 \times 240 }$

$\sf{s= 2390.4m}$



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