Question

A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate
(i) acceleration in m/s2
(ii) distance covered by the bus in metres during this interval.

Answer 1

Acceleration = v - u / t

72km/h = 72 × 5 / 18 m/s = 20m/s

54km/h = 54 × 5 / 18m/s  = 15m/s

    Acceleration = 20-15 / 10 

                          = 5 /10 

                          = 0.5 m/s²

To find distance we use the formula 

             v² = u² + 2aS  

            20² = 15² + 2 × 0.5 × S 

            400 = 225 + 1×S

            400 - 225 = S

          S = 175 m

distance = 175m

Answer 2

Answer :-

Initial velocity, u = 72 km/h = 72 (5/18) m/s = 20 m/s  

Final velocity, v = 54 km/h = 54 (5/18) m/s = 15 m/s  

Time in which velocity changes is, t = 10 s  

(a) Acceleration

Acceleration, a = (v - u)/t  

=> a = (20-15)/10 = 0.5 m/s^2  

(b) Distance travelled by the bus  

v² = u² + 2aS  

20² = 15² + 2 × 0.5 × S  

400 = 225 + 1×S

400 - 225 = S

S = 175 m

Distance = 175 m