Question

Velocity of a particle varies with time as v=t^2-t-2. average speed of the particle in time interval from t=0 to t=4sec?

Answer 1

The average speed is `=4/3ms^-1`

Explanation:

The velocity as a function of time is

`v(t)=t^2-t-2`

The average speed `barv` in the time interval `t in [0,4]` is

`(4-0)barv=int_0^4(t^2-t-2)dt`

`4barv=[t^3/3-t^2/2-2t]_0^4`

`4barv=(4^3/3-4^2/2-2*4)-(0-0-0)`

`4barv=64/3-8-8=16/3`

`barv=4/3ms^-1`

Answer 2

3 m/s

No units for distance were given so I assumed meters.

Explanation:

The vt graph looks like this:

graph{x^2-x-2 [-10, 10, -5, 10]}

You can see that the particle is moving away from the origin at
t = 0 s. It then returns and moves away in a positive direction at
t = 2 s.

The question asks for the average speed which is a scalar quantity. This means we must find the total distance travelled as opposed to the displacement, which is a vector quantity.

To do this I will find the absolute value of the distance travelled between 0 and 2 s then add this to the absolute value of the distance travelled between 2 and 4 s.

In a vt graph the distance travelled is the area between the t axis and the line. We can find this by integration:

For the interval 0 to 2:

`sf(s_1=int_0^2v.dt)`

`sf(v=t^2-t-2)`

`:.` `sf(s_1=int_0^2(t^2-t-2).dt)`

`sf(s_1=[t^3/3-t^2/2-2t]_0^2)`

`sf(s_1=(8/3-4/2-4)-(0))`

`sf(s_1=8/3-6=-10/3color(white)(x)m)`

For the interval 2 to 4:

`sf(s_2=int_2^4(t^2-t-2).dt)`

`sf(s_2=[t^3/3-t^2/2-2t]_2^4)`

`sf(s_2=(64/3-16/2-8)-(-10/3))`

`sf(s_2=16/3-(-10/3)=26/3color(white)(x)m)`

`sf(SigmaS=abs(s_1)+abs(s_2)=26/3+10/3=36/3=12color(white)(x)m)`

This is the total distance travelled.

`:.`average speed = total distance travelled / time taken =12/4 = 3 m/s