Question

Acceleration time graph of particle moving in straight line is given. If initial velocity of particle is 2m/s. Its velocity at $4^{th}$ sec is

Answer 1

Area under a - t graph = change in velocity
= area under trapezium (0-2 sec) + area under the triangle (2-4 sec)
=$(\frac{1}{2}\times 2\times (2+4\left)\right)+\frac{1}{2}\times 2\times 4=v_f-v_i$
$10=v_f-2\Rightarrow v_f=12m/s$