Question

A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer 1

Answer:

Putting all the value, we get

(a) Speed acquired

v = u + at

v = 0 + 0 × 1 × 120

v = 12 m/s

(b) Distance travelled.

(12)² - (0)² = 2(0.1) s

s = 720 m

Hence, Speed acquired finally by the bus is 12 m/s.

And Distance travelled by the bus is 720 m.

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Answer 2

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• Numerical Problem
• A bus starting from rest
• with an acceleration 0.1m/s²
• For 2 minutes

${\sf{\green{\underline{\large{To\:find}}}}}$

• Speed acquired
• Distance Travelel

${\sf{\pink{\underline{\Large{Explanation}}}}}$

☞Firstly change the unit in m/s

• Acceleration=a= 0.1m/s²

• u=0m/s,here the bus start from rest therefore it's initial position is 0

• V=?

• t=2min=120sec

From Equation of Motion

V=u+at

¶utting the value

=>V=u+at

=>V=0+0.1×120

=>V=12m/s

Now for Distance or displacement

• Taking Another Equation of Motion

S=ut + 1/2 at²

V²=u²+2as

You can find the displacement from any of the following Equation

V²=u²+2as

=>(12)²=0+2×0.1S

=>144=o0.2S

=>S=144/0.2

=>S=720m

Hence,S=720m and V=12m/s