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# CBSE Class 9 Answered

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 | 10 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 | 10 Nov, 2021, 10:14: AM

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