CAIE Additional Mathematics Motion

[Maximum number: 10]

In this question, the units are metres and seconds.
A particle P is travelling in a straight line through a fixed point O.
At time t its acceleration, a, is given by $a=(2 t-3)^{2}$ , where $t \geqslant 0$ .
When t=3, P has a velocity of 6.

(a)

Find an expression for the velocity, v, of P at time t.

[ 3 ]

$[\text{Velocity: }]\frac{(2t-3)^3}{3\times 2}+c$
Allow $k(2t-3)^3$ .
$\frac{(2\times 3-3)^3}{6}+c=6$ , so $c=\frac{3}{2}$
M1 for correct calculation to find c.
$v=\frac{(2t-3)^3}{6}+\frac{3}{2}$ oe

(b)

Find the time when P is at rest.

When $t=\frac{5}{2}$ , the displacement of P from O is 4.

[ 3 ]

$\frac{(2t-3)^3}{6}+\frac{3}{2}=0$
M1 FT their expression for v of the form $\frac{(2t-3)^3}{a}+b$ , where $a,b\ne0$ .
$2t-3=\sqrt[3]{-9}$ or $2t-3=-2.08\ldots$ oe
$t=\frac{3+\sqrt[3]{-9}}{2}$ oe isw, or $t=0.460\ldots$

(c)

Find the displacement of P from O when t=3.

[ 4 ]

$[\text{Displacement: }]\frac{(2t-3)^4}{4\times 2\times 6}+\frac{3}{2}t+d$
M1 FT their expression for v of the form $\frac{(2t-3)^3}{a}+b$ , where $a,b\ne0$ .
$\frac{(2\times 2.5-3)^4}{48}+\frac{3}{2}(2.5)+d=4$
M1 correct calculation to find d.
$[\text{Displacement: }]\frac{(2t-3)^4}{48}+\frac{3}{2}t-\frac{1}{12}$
Displacement at t=3: $\frac{293}{48}$ oe isw, or $6.10\ldots$ oe

CAIE IGCSE Additional Math 14.14 Apply Calculus to Kinematics Questions

Exam points

CAIE IGCSE Additional Math 14.14 Apply Calculus to Kinematics Questions question 1

Question (a)

Question (b)

Question (c)