A particle starts from rest at time t=0 and moves in a straight line with an acceleration a(m s^-2). Find the time at which the (a) speed of the particle will be 4 times it's speed at t = 2 s and (b) distance covered by it will be 8 times that travelled by the particle at t= 2 s.
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v=u+at
initially the particle is at rest so u=0
therefore velocity at t=2 is v=a*2
1) now initial velocity is 2a and final velocity is 8a
so for time,
u=0
8a=0+at
therefore t=8
2) apply formula s=ut+(1/2)a*t^2
distance travelled at t=2s
u=0
s=(1/2)a*4
therefore s=2a
so the time taken for the body to travel 8 times the distance travelled at t=2s is
s=ut + (1/2)a*t^2
u=0
so,
16a= (1/2)a*t^2
t^2=32
therefore t=5.66
initially the particle is at rest so u=0
therefore velocity at t=2 is v=a*2
1) now initial velocity is 2a and final velocity is 8a
so for time,
u=0
8a=0+at
therefore t=8
2) apply formula s=ut+(1/2)a*t^2
distance travelled at t=2s
u=0
s=(1/2)a*4
therefore s=2a
so the time taken for the body to travel 8 times the distance travelled at t=2s is
s=ut + (1/2)a*t^2
u=0
so,
16a= (1/2)a*t^2
t^2=32
therefore t=5.66
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