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Question

# A motor car starts from rest and accelerates uniformly for 10 seconds to a velocity of 40 m/s, it then runs at a constant speed and is finally brought to rest in 40 m with a constant acceleration.Total distance covered is 640 meters. Find acceleration, retardation and total time taken.

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Solution

## The journey undertaken is as follows: 1) from A to B.....car starts from rest and accelerates uniformly for 10s until a final velocity of 20m/s. 2)from B to C....runs at a constant speed (20m/s) 3)From C to D.....brought to rest in 40m with a constant acceleration. Use SUVAT equation: where s-dis[lacement, u-initial velocity,v-final velocity, a-acceleration, t-time. V=U+at V2=U2+2as S=ut+1/2at2 S=1/2(U+V)t calculate distance and time taken in each section: i)calculate distance covered between A and B S=1/2(U+V)t =1/2(0+20)10 = 100m Time taken= 10 s ii)distance between B and C: total distance = 640 therefore distance covered is 640-100-40= 500m Time taken=Distance/ speed = 500/20 = 25 seconds iii)distance covered between C and D = 40m (as given)u Time taken = s=1/2(u+v)t therefore t= 2s/(u+v) =2x40/(20+0) =80/20 = 4seconds (4s for stopping) a)Find acceleration: v= u+at here final velocity is 0 therefore at=v where t is total time taken(10s+25+4=39s) a= 0/2 0 acceleration b)retardation: between C and D v=u+at 0=20+4a 4a=-20 a=- 5 therefore retardation = -5m/s2 c)total time taken: that is time between A to D = 10s + 25s + 4s = 39seconds

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