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Question

A particular moves vertically with an upward initial speed v0=10.5 ms1. If its acceleration varies with time as shown in at graph in Fig. find the velocity of the particle at t=4 s.
981458_8ad5e7d6dce94a8eb72bc5b2aaf49117.PNG

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Solution

The velocity of the particle at t=4 s can be given as
v4=v0+Δv
where ΔvA(= area under at) graph during first 4 s)
Referring to at graph (shown in the figure), we have
A=A1+A2A3A4.........(1)
where
A1=5×1=5,A2=12×x×5
A3=12×(1x)×10
A4=12×2×10=10
We can find the value of x as follows:
Using properties of similar triangles, we have x5=1x10
This yields x=13
Substituting x=13 in A2=12×x×5 and A3=12(1x)×10,
A2=56
and A3=103
Then substituting A1,A2,A3 and A4 in Eq. (1) we have
A=7.5Δv=7.5 and v0=10.5 m/s
Hence, we have v4=v0+Δv=10.57.5=3 m/s

1569223_981458_ans_36a5d06ef1ae49379b1944b62a5edf57.PNG

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