Question

a bus accelerates uniformly from 54km/h to 72 km/h in 10 seconds calculate 1.acceleration in m/s 2 2.distance covered by the bus in metres during this interval.

Answer 1

Acceleration = Change in velocity / Time
                     = 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   here v is the final velocitiy (72km/h)
                                              u is the initial velocity(54km/h)
                                               a is the acceleration
                                              S is the distance
             20² = 15² + 2 × 0.5 × S 
             400 = 225 + 1×S
             400 - 225 = S
           S = 175 m
Therefore distance = 175m

Answer 2

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 = 175m