21N.1.SL.TZ0.7

pestleMathematicsAASLPaper 121N· sl-5-5-integration-introduction-areas-between-curve-and-x-axissource ↗

A particle $P$ moves along the $x$ -axis. The velocity of $P$ is $v m s^{- 1}$ at time $t$ seconds, where $v (t) = 4 + 4 t - 3 t^{2}$ for $0 \leq t \leq 3$ . When $t = 0, P$ is at the origin $O$ .

Find the value of $t$ when $P$ reaches its maximum velocity.

[2]

a.i.

Show that the distance of $P$ from $O$ at this time is $\frac{88}{27}$ metres.

[5]

a.ii.

Sketch a graph of $v$ against $t$ , clearly showing any points of intersection with the axes.

[4]

Find the total distance travelled by $P$ .

[5]

Markscheme / solution

valid approach to find turning point ( $v' = 0, - \frac{b}{2 a}$ , average of roots) (M1)

$4 - 6 t = 0$ OR $- \frac{4}{2 (- 3)}$ OR $\frac{- \frac{2}{3} + 2}{2}$

$t = \frac{2}{3}$ (s) A1

[2 marks]

a.i.

attempt to integrate $v$ (M1)

$\int v d t = \int (4 + 4 t - 3 t^{2}) d t = 4 t + 2 t^{2} - t^{3} (+ c)$ A1A1

Note: Award A1 for $4 t + 2 t^{2}$ , A1 for $- t^{3}$ .

attempt to substitute their $t$ into their solution for the integral (M1)

distance $= 4 (\frac{2}{3}) + 2 {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{3}$

$= \frac{8}{3} + \frac{8}{9} - \frac{8}{27}$ (or equivalent) A1

$= \frac{88}{27}$ (m) AG

[5 marks]

a.ii.

valid approach to solve $4 + 4 t - 3 t^{2} = 0$ (may be seen in part (a)) (M1)

$(2 - t) (2 + 3 t)$ OR $\frac{- 4 \pm \sqrt{16 + 48}}{- 6}$

correct $x$ - intercept on the graph at $t = 2$ A1

Note: The following two A marks may only be awarded if the shape is a concave down parabola. These two marks are independent of each other and the (M1).

correct domain from $0$ to $3$ starting at $(0, 4)$ A1

Note: The $3$ must be clearly indicated.

vertex in approximately correct place for $t = \frac{2}{3}$ and $v > 4$ A1

[4 marks]

recognising to integrate between $0$ and $2$ , or $2$ and $3$ OR $\int_{0}^{3} |4 + 4 t - 3 t^{2}| d t$ (M1)

$\int_{0}^{2} (4 + 4 t - 3 t^{2}) d t$

$= 8$ A1

$\int_{2}^{3} (4 + 4 t - 3 t^{2}) d t$

$= - 5$ A1

valid approach to sum the two areas (seen anywhere) (M1)

$\int_{0}^{2} v d t - \int_{2}^{3} v d t$ OR $\int_{0}^{2} v d t + |\int_{2}^{3} v d t|$

total distance travelled $= 13$ (m) A1

[5 marks]

Examiners’ report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]