Question

Acceleration-time graph of a particle moving in a straight line is as shown in figure. At time `t = 0,` velocity of the particle is zero. Find

(a) average acceleration in a time interval from `t = 6 s` to `t = 12 s,`
(b) velocity of the particle at `t = 14 s.`

Answer 1

Correct Answer - A::B
(a) `a_(av)=(v_f-v_i)/t=(v_12-v_6)/(12-6)`
`=((-10)-(20))/6=-5m//s^2`
(b)`Deltav=v_f-v_i`=net area of a-t graph
`:. v_(14)-v_0=40+30+40-20=90`
But `v_0=0`
`:. v_(14)=90m//s`