A particle moves vertically with an upward initial speed v0 = 10.5 ms-1. If its acceleration varies with time as shown in a-t graph in figure, find the velocity of the particle at t = 4 s.

The velocity of the particle at t = 4 s can be given as v→4=v→0+Δv→  ……..(i)where Δv→≡A (= area under a-t graph during first 4 s) Referring to a-t graph (shown in figure), we have             A=A1+A2−A3−A4       .......(ii)where A1=5×1=5,A2=12×x×5            A3=12×(1−x)×10 and A4=12×2×10=10We can find the value of x as follows: Using properties of similar triangles, we have x5=1−x10This yields x=13Substituting x=13 in A2=12×x×5 and A3=12(1−x)×10we have A2=5/6 and A3=10/3Then substituting A1,A2,A3 and A4 in Eq. (ii), we have A=−7.5⇒Δv=−7.5 and v→0=10.5ms−1 Hence, we have v4=v0+Δv=10.5−7.5=3ms−1.