Question

A particle moves along straight line and its velocity depends on time as v=3t - t^2 v in m/s t in second. Find the average velocity for first 5 second

Answer 1

Answer:

The average velocity is 0.834 m/s.

Explanation:

Given that,

Velocity $v = 3t-t^2$

The velocity is the first derivative of position of the particle.

$v=\dfrac{dx}{dt}=3t-t^2$

On integrating both side

$\int{\dfrac{dx}{dt}}=\int{3t-t^2}$

$\int{dx}=\int{3t-t^2}dt$

$x=\dfrac{3t^2}{2}-\dfrac{t^3}{3}+C$....(I)

here, x = 0, t = 0 then C = 0

Now, Put the value of C in equation (I)

$x=\dfrac{3t^2}{2}-\dfrac{t^3}{3}$....(II)

Now, put the value of t in equation (II)

$x=\dfrac{3\times25}{2}-\dfrac{5^3}{3}$

$x=-4.17\ m$

Average velocity :

The average velocity is equal to the total distance divided by the total time.

Formula of average velocity is defined as,

$v=\dfrac{D}{T}$

Where, D = total distance

T = total time,

Put the value into the formula,

$v_{avg}=\dfrac{-4.17}{5}$

$v_{avg}=-0.834\ m/s$

Negative sign shows the opposite direction of motion.

Hence, The average velocity is 0.834 m/s.