19M.2.AHL.TZ2.H_6

pestleMathematicsAAHLPaper 219M· sl-5-9-kinematics-problemssource ↗

A particle moves along a horizontal line such that at time $t$ seconds, $t$ ≥ 0, its acceleration $a$ is given by $a$ = 2 $t$ − 1. When $t$ = 6 , its displacement $s$ from a fixed origin O is 18.25 m. When $t$ = 15, its displacement from O is 922.75 m. Find an expression for $s$ in terms of $t$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to integrate $a$ to find $v$ M1

$v = \int {a\,{\text{d}}t = \int {\left( {2t - 1} \right)} } \,{\text{d}}t$

$= {t^2} - t + c$ A1

$s = \int {v\,{\text{d}}t = \int {\left( {{t^2} - t + c} \right)} } \,{\text{d}}t$

$= \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + ct + d$ A1

attempt at substitution of given values (M1)

at $t = 6{\text{,}}\,\,\,18.25 = 72 - 18 + 6c + d$

at $t = 15{\text{,}}\,\,\,922.75 = 1125 - 112.5 + 15c + d$

solve simultaneously: (M1)

$c = - 6{\text{,}}\,\,d = 0.25$ A1

$\Rightarrow s = \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + - 6t + \frac{1}{4}$

[6 marks]

Examiners’ report

[N/A]