Question

A car starts from rest and accelerates uniformly for
10 second. 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 deceleration. Tolal distance covered is
640m. Find acceleration, retardation & total time taken.
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Answer 1

$\bf\large\underline\green{Solution:-}$

❒________________

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.

Answer 2

Answer is given below