Question

The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t2-4.0t3m. Find (a) the displacement, average velocity, average acceleration between 0.0 s and 1.0,

Answer 1

`\Delta x=1-0=1\ m`

`\Delta v=-2-0=-2\ m.s^(-1)`

`\Delta a=-14-10=-24\ m.s^(-1)`

Step-by-step explanation:

The equation governing the position of the particle moving along x-axis is given as:

`x=5* t^2-4* t^3`

we know that the time derivative of position gives us the velocity:

`(d)/(dt) x=v`

`v=10\ t-12\ t^2`

and the time derivative of of velocity gives us the acceleration:

`(d)/(dt) v=a`

`a=10-24\ t`

Now, when t = 0

`x=0\ m`

`v=0\ m.s^(-1)`

`a=10\ m.s^(-2)`

When t=1 s

`x_1=5* 1^2-4* 1^3=1\ m`

`v_1=10* 1-12* 1^2=-2\ m.s^(-1)`

`a_1=10-24* 1=-14\ m.s^(-2)`

Hence,

Displacement between the stipulated time:

`\Delta x=x_1-x`

`\Delta x=1-0=1\ m`

Velocity between the stipulated time:

`\Delta v=v_1-v`

`\Delta v=-2-0=-2\ m.s^(-1)`

Acceleration between the stipulated time:

`\Delta a=a_1-a`

`\Delta a=-14-10=-24\ m.s^(-1)`

Here negative sign indicates that the vectors are in negative x direction.