Question

A bus accelerates uniformly 54 km/h to 72 km/h in 10 sec calculate acceleration in m/s² and distance covered by the bus in meters 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