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.

Solution

(a) Given, the bus 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=(v-u)/t

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

= (0.1 m.s^-2 * 120 s) + 0 m.s^-1

= 12 m.s^-1 + 0 m.s^-1

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

(b) As per the third motion equation, 2as = v² – u²

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

Distance, s =(v² – u²)/2a

=(12² – 0²)/2(0.1)

Therefore, s = 720 m.

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