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.

Asked by Vishal kumar | 2 years ago |  364

##### Solution :-

(a) Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) $$=0.1\,m.s^{-2}$$

Time = 2 minutes = 120 s

Acceleration is given by the equation $$a=\frac{v-u}{t}$$

Therefore, terminal velocity (v) = (at) + u

$$=(0.1\,m.s^{-2}\times 120s)$$ $$+0\,m.s^{-1}$$

= 12m.s-1 + 0 m.s-1

Therefore, terminal velocity (v) = 12m/s

(b) As per the third motion equation, 2as = $$v^2-u^2$$

Since a = 0.1 m.s-2 , v = 12 m.s-1 , u = 0 m.s-1 , and t = 120s, the following value for s (distance) can be obtained.

Distance, $$s=\frac{v^2-u^2}{2a}$$

$$=\frac{12^2-0^2}{2(0.1)}$$

Therefore, s = 720m.

The speed acquired is 12m.s-1 and the total distance traveled is 720m

Answered by Vishal kumar | 2 years ago

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