A particular moves vertically with an upward initial speed v0=10.5ms−1. If its acceleration varies with time as shown in a−t graph in Fig. find the velocity of the particle at t=4s.
Open in App
Solution
The velocity of the particle at t=4s can be given as →v4=→v0+Δ→v where Δ→v≡A(= area under a−t) graph during first 4s) Referring to a−t graph (shown in the figure), we have A=A1+A2−A3−A4.........(1) where
A1=5×1=5,A2=12×x×5 A3=12×(1−x)×10
A4=12×2×10=10 We can find the value of x as follows: Using properties of similar triangles, we have x5=1−x10 This yields x=13 Substituting x=13 in A2=12×x×5 and A3=12(1−x)×10, A2=56
and A3=103 Then substituting A1,A2,A3 and A4 in Eq. (1) we have A=−7.5⇒Δv=−7.5 and →v0=10.5m/s Hence, we have v4=v0+Δv=10.5−7.5=3m/s