Starting from rest, an object is uniformly accelerated to a speed of 60m/s in 20s. The object travels with this speed for next 40s and is then brought to rest by a uniform retardation in the next 30s.Sketch velocity time graph and calculate the acceleration, retardation and the total distance travelled.
Hint: acceleration =rate of change of velocity. distance covered =area under velocity-time graph .
Asked by chaitanyas5425 | 10th Nov, 2021, 09:16: AM
Acceleration during initial phase of travel is determined from the following equation of motion
v = u + ( a t )
where u = initial speed = 0 and v = 60 m/s is speed attained with uniform acceleration a in t = 20 s
a = v / t = 60 / 20 = 3 m/s2
Distance S1 travelled in initial phase , S1 = (1/2) a t2 = 0.5 × 3 × 20 × 20 = 600 m
Distance S2 travelled in uniform speed 60 m/s for 40 s is calculated as
S2 = 60 × 40 = 2400 m
Retardation during last phase of travel is determined from the following equation of motion
v = u - ( ar t )
where u = initial speed = 60 m/s and v = 0 m/s is final speed after t = 30 s
ar = u / t = 60 / 30 = 2 m/s2
Distance S3 travelled while retardation is calculated as follows
S3 = ( u t ) - [ (1/2) a t2 ] = ( 60 × 30 ) - ( 0.5 × 2 × 30 × 30 ) = 900 m
Total distance S = S1 + S2 + S3 = 600 + 2400 + 900 = 3900 m
Answered by Thiyagarajan K | 10th Nov, 2021, 10:14: AM
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