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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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