A bus accelerates uniformly from 54 km/h to 72 km/h in 10 s. Calculate acceleration and distance covered.
Answers
Answered by
56
acceleration is taken out by law v=u + at
72 k/hr= 20m/s
54 k/hr=15m/s
20=15 + 10a
-10a= -5
a= -5/-10
a=0.5 m/s^2
Distance covered is to be found by 2as= v^2-u^2
2X0.5s= (20)^2 -( 15)^2
s=400 - 225
s=175 m
72 k/hr= 20m/s
54 k/hr=15m/s
20=15 + 10a
-10a= -5
a= -5/-10
a=0.5 m/s^2
Distance covered is to be found by 2as= v^2-u^2
2X0.5s= (20)^2 -( 15)^2
s=400 - 225
s=175 m
Answered by
50
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
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