CAIE IGCSE Additional Math 14.14 calculus kinematics

Use this calculus-kinematics page to move between s, v and a, solve for times and interpret distance travelled or maximum displacement.

Exam points

Differentiate displacement to find velocity or acceleration for particle motion.
Integrate acceleration or velocity and use conditions to find missing constants.
Solve for rest times, maximum displacement or total distance over a time interval.

[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 ]

$[\text{Velocity: }]\frac{(2t-3)^3}{3\times 2}+c$
M1
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
A1

(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
A1
$t=\frac{3+\sqrt[3]{-9}}{2}$ oe isw, or $t=0.460\ldots$
A1

(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.
$[\text{Displacement: }]\frac{(2t-3)^4}{48}+\frac{3}{2}t-\frac{1}{12}$
A1
Displacement at t=3: $\frac{293}{48}$ oe isw, or $6.10\ldots$ oe
A1

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14.14. Apply calculus to kinematics question 1

Question (a)

Question (b)

Question (c)