21M.2.AHL.TZ2.6

pestleMathematicsAIHLPaper 221M· ahl-5-13-kinematic-problemssource ↗

A particle $P$ moves along the $x$ -axis. The velocity of $P$ is $v {m s}^{- 1}$ at time $t$ seconds, where $v = - 2 t^{2} + 16 t - 24$ for $t \geq 0$ .

Find the times when $P$ is at instantaneous rest.

[2]

Find the magnitude of the particle’s acceleration at $6$ seconds.

[4]

Find the greatest speed of $P$ in the interval $0 \leq t \leq 6$ .

[2]

The particle starts from the origin $O$ . Find an expression for the displacement of $P$ from $O$ at time $t$ seconds.

[4]

Find the total distance travelled by $P$ in the interval $0 \leq t \leq 4$ .

[3]

Markscheme / solution

solving $v = 0$ M1

$t = 2, t = 6$ A1

[2 marks]

use of power rule (M1)

$\frac{d v}{d t} = - 4 t + 16$ (A1)

$(t = 6)$

$\Rightarrow a = - 8$ (A1)

magnitude $= 8 {m s}^{- 2}$ A1

[4 marks]

using a sketch graph of $v$ (M1)

$24 {m s}^{- 1}$ A1

[2 marks]

METHOD ONE

$x = \int v d t$

attempt at integration of $v$ (M1)

$- \frac{2 t^{3}}{3} + 8 t^{2} - 24 t (+ c)$ A1

attempt to find $c$ (use of $t = 0, x = 0$ ) (M1)

$c = 0$ A1

$(x = - \frac{2 t^{3}}{3} + 8 t^{2} - 24 t)$

METHOD TWO

$x = \int_{0}^{t} v d t$

attempt at integration of $v$ (M1)

${[- \frac{2 t^{3}}{3} + 8 t^{2} - 24 t]}_{0}^{t}$ A1

attempt to substituted limits into their integral (M1)

$x = - \frac{2 t^{3}}{3} + 8 t^{2} - 24 t$ A1

[4 marks]

$\int_{0}^{4} |v| d t$ (M1)(A1)

Note: Award M1 for using the absolute value of $v$ , or separating into two integrals, A1 for the correct expression.

$= 32 m$ A1

[3 marks]

Examiners’ report

[N/A]