M2 2017 4 Edexcel

4

(a)

The particle is instantaneously at rest when v = 0, ie. when:

v = 3t^2 - 16t + 21 = (3t - 7)(t - 3) = 0

So either t = \dfrac 7 3 or t = 3. Since \dfrac 7 3 < 3, we have t_1 = \dfrac 7 3 and t_2 = 3.

(b)

We know that the acceleration a is given by:

\displaystyle a = \frac {\mathrm dv} {\mathrm dt}

so that:

a = 6t - 16

At t = \dfrac 7 3 we have:

a = 6 \left(\dfrac 7 3\right) - 16 = -2

So the magnitude of the acceleration of P at t_1 is 2 \text{ms}^{-1}.

©

Note that the displacement s of P is given by:

\displaystyle s = \int v \mathrm dt

Since v \le 0 for t_1 \le t \le t_2, P does not change direction so we the distance it travels in this time is given by:

\displaystyle \left|\int_{\frac 7 3}^3 (3t^2 - 16t + 21) \mathrm dt\right|

We have:

\begin{align*}\int_{\frac 7 3}^3 (3t^2 - 16t + 21) \mathrm dt & = \left[t^3 - 8t^2 + 21 t\right]_{\frac 7 3}^3 \\ & = 3^3 - 8 \times 3^2 + 21 \times 3 - \left(\frac 7 3\right)^3 + 8 \left(\frac 7 3\right)^2 - 21 \times \left(\frac 7 3\right) \\ & = -\frac 4 {27}\end{align*}

So the distance travelled is \dfrac 4 {27} \text m = 0.15 \text m \text { (2sf)}.

(d)

P returns to O at time t only if:

t^3 - 8t^2 + 21t = 0

for t \ne 0. (since P starts at O) That is:

t^2 - 8t + 21 = 0

However, this has discriminant:

8^2 - 4 \times 21 = -20 < 0

so the equation has no real solutions in t, so P cannot return to O.

Sweet :candy: