A car accelerates with uniform rate from rest on a straight road. the distance travelled in the last second of a three second interval from the start is 15m then find the distance travelled in first second?
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Lets assume that it is accelerating to with a m/s^2
And initial velocity (u) =0
At t=2 and t=3
S=ut+0.5at^2
S(2)=0.5a2^2=2a
S(3)=0.5a3^2=4.5a
Distance travelled in 3rd second is
D=s(3)-s(2)=2.5a=15m
Hence a=6m/s^2
Thus distance in first second
D(t=1)=0.5at^2=0.5*6*1=3m
And initial velocity (u) =0
At t=2 and t=3
S=ut+0.5at^2
S(2)=0.5a2^2=2a
S(3)=0.5a3^2=4.5a
Distance travelled in 3rd second is
D=s(3)-s(2)=2.5a=15m
Hence a=6m/s^2
Thus distance in first second
D(t=1)=0.5at^2=0.5*6*1=3m
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