Starting from rest, acceleration of a particle is $a = 2(t – 1)$. The velocity of the particle at $t = 5s$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 15 m/s
$\frac{dv}{dt}=2(t-1)$
or $\int_0^vdv=\int_0^52(t-1)dt$
or $v=2\left|\left(\frac{t^2}{2}-t\right)\right|_0^5=2\left(\frac{5^2}{5}-t\right)=15m/s$