Question

The initial velocity of a particle is x m/s (at t=0) and acceleration a is function for time, given by, a=6t. Which of the following relation is correct for final velocity y after time t?

Answer 1

Answer:

$y=x+3t^2$

Explanation:

1) We have,

Initial velocity , u =x m/s

Final velocity , v =y m/s

acceleration ,a = 6t

2) Then , we know that:

After time t,

$a=6t\\ \\=>\frac{dv}{dt}=6t\\ \\=>dv=6tdt\\ \\=>\int\limits^y_x {dv} \,=\int\limits^t_0 {6t} \, dt\\ \\=>y-x=\frac{6t^2}{2}=3t^2\\ \\=>y=x+3t^2$

Hence, after time 't' , relation for final velocity 'y' is given by

$\boxed{y=x+3t^2}$

Answer 2

Answer: y = x +3t^2

Explanation: