Question

(b) A train staring from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find out the
(i) acceleration
(ii) the distance travelled by the train for attending this velocity?​

Answer 1

Explanation:

0.24 velocity travelled by the train

acceleration does not change

Answer 2

$\huge\sf { \dag Question: }$

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find out the

(i) acceleration

(ii) the distance travelled by the train for attending this velocity?

$\huge\sf { \dag Solution:}$

${\underline {\underline {\rm {\red { \huge Given }}}}}$

• $\sf\pink { u ( initial \: velocity ) = 0}$

$\sf { [As \: it \: started \: from \: rest ] }$

• $\sf\pink { v ( final \: velocity ) = 72 km/h= 20m/s}$

$\sf \blue { Converting \: to \: its \: standard \: form }$

$\sf { ☞ \cancel {{72}}^{ \: \: 4} \times \dfrac {5}{\cancel {18}} }$

$\sf { ☞ 4 \times 5 = 20m/s}$

• $\sf\pink { t ( time \: taken ) = 5minutes = 300 s}$

$\sf \blue { Converting \: to \: its \: standard \: form }$

$\sf { ☞ 5 \times 60 = 300 seconds }$

${\underline {\underline {\rm {\red { \huge To \: find: }}}}}$

• $\sf { acceleration }$
• $\sf { total \: distance \: travelled }$

${\underline {\underline {\rm {\red { \huge Calculation: }}}}}$

$\sf\purple { To \: find \: acceleration :}$

$\tt\green { a = \dfrac{v-u}{t} }$

$\sf { :\implies a = \dfrac{20-0}{300} }$

$\sf { :\implies a = \dfrac{2\cancel0}{30\cancel0} }$

$\sf { :\implies a = \dfrac{\cancel2}{\cancel {30}}= \dfrac{1}{15}m/{s}^{2} }$

$\sf {\therefore Acceleration \: of \: the \: train \: was \: \dfrac{1}{15}m/{s}^{2} }$

__ __ __ __ __ __ __

$\sf\purple { To \: find \: total \: distance \: travelled:}$

by using second equation of kinematics

$\tt\green { s =ut + \dfrac{1}{2}a{t}^{2} }$

$\sf { :\implies s = (0 \times 300 )+ \dfrac{1}{2} \times \dfrac{1}{15} \times 300 \times 300 }$

$\sf { :\implies s = \dfrac{1}{2} \times \dfrac{1}{\cancel {15}} \times {\cancel{300}}^{ \: \: \: 20 } \times 300 }$

$\sf { :\implies s = \dfrac{1}{2} \times 20 \times 300 }$

$\sf { :\implies s = \dfrac{1}{\cancel 2} \times \cancel {6000} = 3000m }$

$\sf { In \: km- }$

$\sf { 1000m = 1km }$

$\sf { so, 3000m = 3km }$

$\sf { \therefore Distance \: travelled \: by \: the \: train \: was 3km \: or 3000 m }$

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${\underline {\underline {\rm {\red { \huge Required \: Answers: }}}}}$

• $\sf { \dag Acceleration = \dfrac{1}{15}m/{s}^{2} }$

• $\sf { \dag Distance = 3km \: or 3000 m }$

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