A car starts from rest, If the velocity of the car becomes 650m/s after 25 second calculate the acceleration of the car.
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650-0/25 = 26m/s²........
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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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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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