Question

A train starting from railway station and moving with uniform acceleration attains a speed 40km/hr in 10min,find its acceleration

Answer 1

Answer :-

0.0185m/s^2

Explanation :-

Given :

Initial velocity,u = 0

Final velocity,v = 40km/hr = 11.11ms [1km/hr = 1000/(1*60*60)]

Time,t = 10 minutes = 600s  [1min = 60s]

To Find :

Acceleration,a = ?

Solution :

We know that,

$\sf{}v=u+at$

So,

$\implies \sf{}40km/hr=0km/hr+a\times 10$

$\implies \sf{}11.11m/s=0m/s+a\times 600s$

$\sf{}\implies 11.11m/s=a\times 600s$

$\implies \sf{}\dfrac{11.11m/s}{600s}=a$

$\therefore \sf{}a=0.0185m/s^2$

Therefore,it acceleration is equal to 0.0185m/s^2

☆Know more☆

$\sf{}Second\ equation\ of\ motion: s=ut+\dfrac{1}{2}at^2$

$\sf{}Third\ equation\ of\ motion: v^2=u^2+2as$

Answer 2

Given:

• Speed= 40km/ hr
• Time= 10min
• Initial velocity = 0m/s

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Need to find:

Acceleration =?

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We know,

1km= 1000m

1hour= 3600 sec

Speed:

$= \dfrac{40km}{1hr}$

$= \dfrac{40 \times 5 \: m}{18 \: sec}$

$= \dfrac{100}{9}$

=$\red{11.1m/s}$

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1min = 60sec

10min = 60×10= 600sec

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• v= 11.11 m/s
• u= 0m/s
• t= 600sec

We know,

First equation of motion:

$v = u + at$

$\implies \: 11. 11 = 0 + a \times 600$

$\implies \: 11.11 = 600a$

$\implies \: a = 11.11 \div 600$

$\green{ \implies \: a = 0.0185 \dfrac{m}{ {sec}^{2} }}$