18N.2.SL.TZ0.S_4

pestleMathematicsAASLPaper 218N· sl-5-9-kinematics-problemssource ↗

A particle moves along a straight line so that its velocity, $v$ m s⁻¹, after $t$ seconds is given by $v\left( t \right) = {1.4^t} - 2.7$ , for 0 ≤ $t$ ≤ 5.

Find when the particle is at rest.

[2]

Find the acceleration of the particle when $t = 2$ .

[2]

Find the total distance travelled by the particle.

[3]

Markscheme / solution

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

valid approach (M1)

eg $v\left( t \right) = 0$ , sketch of graph

2.95195

$t = {\text{lo}}{{\text{g}}_{1.4}}2.7$ (exact), $t = 2.95$ (s) A1 N2

[2 marks]

valid approach (M1)

eg $a\left( t \right) = v'\left( t \right)$ , $v'\left( 2 \right)$

0.659485

$a\left( 2 \right)$ = 1.96 ln 1.4 (exact), $a\left( 2 \right)$ = 0.659 (m s⁻²) A1 N2

[2 marks]

correct approach (A1)

eg $\int_0^5 {\left| {v\left( t \right)} \right|} \,{\text{d}}t$ , $\int_0^{2.95} {\left( { - v\left( t \right)} \right)} \,{\text{d}}t + \int_{295}^5 {v\left( t \right)} \,{\text{d}}t$

5.3479

distance = 5.35 (m) A2 N3

[3 marks]

Examiners’ report

[N/A]