Given, initial speed of the bus, u = 0, acceleration, a=0.1 ms−2 and time taken, t=2 minutes=2×60=120 s
Let v be the final velocity and s be the distance covered.
(a) From the first equation of motion, v = u + at,
v=0+0.1×120
⇒ v=12 ms−1
(b) According to the third equation of motion, v2−u2=2as
(12)2−(0)2=2×0.1×s
⇒s=720 m
Speed acquired by the bus is 12 ms−1 and the distance travelled is 720 m.