a body is moving with uniform acceleration . its velocity after 5 sec is 25m/s and after 8 sec is 34m/s . calculate the distance it will cover in the 10th second?
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at t=5 sec initial velocity u = 25m/s
at t=8 sec final velocity v =34m/s
the time interval for which the car travelled = 8-5 = t = 3 sec
therefore the constant acceleration a is given by
a= (v-u)/t
a=(34-25)/3
a=3m/s^2
therefore initial velocity at t=0
u1=v-at
u1=34-3(8)
u1=10m/s
the distance it will cover in the tenth seconds is equal to
(the distance covered after 9 secs) - ( the distance covered after 10 secs)
we use the formula s=ut+1/2 at^2
therfore distance covered after 9 secs = 10(9) + 1/2(3)(9)^2=211.5m
similarly distance covered after 10 sec = 10(10) + 1/2(3)(10)^2=250m
therefore distance covered in the tenth second=250-211.5=38.5m
at t=8 sec final velocity v =34m/s
the time interval for which the car travelled = 8-5 = t = 3 sec
therefore the constant acceleration a is given by
a= (v-u)/t
a=(34-25)/3
a=3m/s^2
therefore initial velocity at t=0
u1=v-at
u1=34-3(8)
u1=10m/s
the distance it will cover in the tenth seconds is equal to
(the distance covered after 9 secs) - ( the distance covered after 10 secs)
we use the formula s=ut+1/2 at^2
therfore distance covered after 9 secs = 10(9) + 1/2(3)(9)^2=211.5m
similarly distance covered after 10 sec = 10(10) + 1/2(3)(10)^2=250m
therefore distance covered in the tenth second=250-211.5=38.5m
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