A particle moves vertically with an upwards initial speed `v_0 = 10.5 ms^(-1).` If its acceleration varies with time as shown in a-t graph in figure, find the velocity of the particles at t = 4s.

Answer

3

Explanation

The velocity of the particle at
t = 4 s can be given as
`vecv_4 = vecv_0 + Deltavecv` ……..(i)
where `Deltavecv equivA `(= area under a-t
graph during first 4 s)
Referring to a -t graph (shown in
figure), we have
`A = A_1+A_2 - A_3-A_4`
where
`A_1 = 5 xx 1 = 5, A_2 = 1/2 xx x xx 5`
`A_3 = 1/2 xx (1-x) xx 10 and A_4 = 1/2 xx 2 xx 10 = 10`
We can find the value of x as follows:
Using properties of similar triangles, we have `x/5 = (1-x)/10.`
This yields ` x = 1/3`
Substituting `x = 1/3 "in" A_2 = 1/2 xx x xx 5 and A_3 = 1/2 (1-x) xx`
10, we have `A_2 = 5//6 `and `A_3 = 10//3`.
Then substituting `A_1,A_2,A_3`and `A_4` in Eq. (ii), we have
`A = -7.5 rArr Deltav = -7.5 and vecv_0 = 10.5 ms^(-1)`.
hence, we have `v_4 = v_0 + Deltav = 10.5 - 7.5 = 3 ms^(-1)`.