Question

a particle position as a function of time is described as y(t)=2t^2+3t^2+4. what is the avarage velocity of the particle t=0to t=3sec

Answer 1

average v = total displacement / total time
v at t=0 .. 4m/s
v at 3 sec ... 22 m/s
so delta v = 18
total t=3
hence av. V = 6m/s

Answer 2

Answer:

Velocity, v = 30 m/s

Explanation:

It is given that,

The position of a particle as a function of time is given by :

$y(t)=2t^2+3t^2+4$

We know that, the velocity of a particle is given by :

$v=\dfrac{dy}{dt}$

$v=\dfrac{d(2t^2+3t^2+4)}{dt}$

$v=4t+6t$

Velocity at t = 0, v = 0

Velocity at t = 3 s, $v=4(3)+6(3)=30\ m/s$

So, the average velocity of the particle t=0 to t=3 sec is 30 m/s. Hence, this is the required solution.