Question

A bus starts from rest moves with uniform acceleration of 0.1 for 2 min.Find 1 Speed acquired 2. the distance travelled

Answer 1

Here acceleration (a) = 0.1 ms
Time (t) = 2 min.=2*60=120 s
Initial speed u = 0
Speed acquired
From I st equation of motion, speed acquired v=u+at=0+0.1*120+12 ms
Distance travelled
From II st equation of motion, Distance travelled,
s=ut+1/2at2=0*1201/2*0.1*120*120
=0.1*60.120=720 m.

Answer 2

Hi there!

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Given ;

Initial velocity, u = 0.

Acceleration, a = 0.1 m/s²

Time taken, t = 2 mins = 120 secs.

Solution ;

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(a) Using first equation of motion we have ;

v = u + at

⇒ v = 0 + 0.1 × 120

⇒ v = 12 m/s.

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(b) Using third equation of motion, we have ;

v² - u² = 2as

⇒ ( 12 )² - (0)² = 2 × 0.1 × s

⇒ 144 = 0.2 s

⇒ s = 1440/ 2

⇒ s = 720 m.

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Hence,

Speed acquired by the bus is 12 m/s.

Distance travelled by the bus is 720m.

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Thanks for the question !

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