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CBSE Class 9 Answered

An object starts from rest and is uniformly accelerated so that its speed is 60 m/s after 20s. If it travels with this speed for 40 s and is then brought to rest by a uniform retardation in 30 s. Sketch the velocity-time graph and calculate the acceleration, the retardation and the total distance travelled.
Asked by sbajaj544 | 05 Jun, 2018, 11:34: AM
answered-by-expert Expert Answer
Object starts from rest ( initital velocity u =0),  accelerated and speed v becomes 60 m/s after 20 s.
 
v = u+a×t = 0+a×20 ; hence acceleration a = 3 m/s2
 
distance travelled at initial 20 s with acceleration,  S1 = (1/2)×a×t2 = (1/2)×3×400 = 600 m
 
distance travelled with constant speed 60 m/s for 40 s, S2  = 60×40 = 2400 m.
 
retardation given to reduce the speed from 60 m/s to 0 m/s in 30 s;
 
hence retardation  is obtained from  v= u-a×t , here v =0, u = 60 m/s, t = 30 s;
 
hence 0 = 60-a×30 ; hence a = 2 m/s2
 
distance travelled at final 30 s with retardation,  S3 = u×t+(1/2)×a×t2 = 60×30-(1/2)×2×900 = 900 m
 
total distance travelled S1+S2+S3 = 600+2400+900 = 3900 m


Answered by Thiyagarajan K | 05 Jun, 2018, 02:18: PM
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