Question

I can not do this nor understand?

Answer 1

Using calculus, take the integral of velocity to get the equation for displacement at any time, then subtract the value at `t = 6` from that at `t = 8` to get `-9` meters.

Explanation:

Remember the definition of velocity being the derivative of displacement with respect to time:

`v = (ds)/(dt)`

Taking the integral with respect to time, we should get the displacement:

`int v * dt = int (ds)/(dt) * dt`

`s = int v * dt`

So, to get the displacement at any time, solve this integral by substituting for the velocity:

`v = 6 - (3t)/2 rarr v = -3/2 t + 6`

`s = int (-3/2 t + 6) * dt`

Split the integral:

`s = int (-3/2 t) * dt + int (6) * dt`

Take out the constants (by applying linearity):

`s = -3/2 int (t) * dt + 6 int (1) * dt`

Then solve for each integral separately:

`s = -3/2 * (t^2)/2 + 6 * t`

And simplify:

`s = -3/4 t^2 + 6t`

Since we're looking for the change in displacement from `t = 6` to `t = 8`, simply solve for the displacement at each point in time, then subtract:

`Deltas = s_8 - s_6 = (-3/4 (8)^2 + 6(8)) - (-3/4 (6)^2 + 6(6))`

`= (-3/4 * 64 + 6 * 8) - (-3/4 * 36 + 6 * 6)`

`= (-3 * 16 + 48) - (-3 * 9 + 36)`

`= (-48 + 48) - (-27 + 36) = -(36 + 27) = -9 m`

Therefore, the change in displacement between `t = 6` and `t = 8` is `-9` meters.

Answer 2

Given is

`v=6-(3t)/2` for `(6<=t<=8)` ......(1)

Comparing with kinematic expression

`v=u+at`, we get
`u=6ms^-1` and
`a=-3/2ms^-2`

For displacement we have the kinematic expression

`s=s_0+ut+1/2at^2` ......(2)

We are required to find change in displacement for `t=6` to `t=8`.

Taking initial time at `t=6=>s_0=0`. We need initial velocity `u` at `t=6`. From (1) we get

`v(6)=6-(3xx6)/2=-3m^-1`

Duration for which change in displacement is to be calculated `=8-6=2s`.

Therefore from (2) we get

`Deltas=s(8)-s(6)=(-3)xx2+1/2(-3/2)2^2`
`=>Deltas=-6-3=-9m`