22M.2.SL.TZ2.6

pestleMathematicsAASLPaper 222M· sl-5-9-kinematics-problems, sl-5-6-differentiating-polynomials-n-e-q-chain-product-and-quotient-rulessource ↗

A particle moves in a straight line such that its velocity, $v m s^{- 1}$ , at time $t$ seconds is given by $v = \frac{(t^{2} + 1) \cos t}{4}, 0 \leq t \leq 3$ .

Determine when the particle changes its direction of motion.

[2]

Find the times when the particle’s acceleration is $- 1.9 {m s}^{- 2}$ .

[3]

Find the particle’s acceleration when its speed is at its greatest.

[2]

Markscheme / solution

recognises the need to find the value of $t$ when $v = 0$ (M1)

$t = 1.57079 \dots (= \frac{π}{2})$

$t = 1.57 (= \frac{π}{2})$ (s) A1

[2 marks]

recognises that $a (t) = v' (t)$ (M1)

$t_{1} = 2.26277 \dots, t_{2} = 2.95736 \dots$

$t_{1} = 2.26, t_{2} = 2.96$ (s) A1A1

Note: Award M1A1A0 if the two correct answers are given with additional values outside $0 \leq t \leq 3$ .

[3 marks]

speed is greatest at $t = 3$ (A1)

$a = - 1.83778 \dots$

$a = - 1.84 ({m s}^{- 2})$ A1

[2 marks]

Examiners’ report

In part (a) many did not realize the change of motion occurred when $v = 0$ . A common error was finding $v (0)$ or thinking that it was at the maximum of $v$ .

In part (b), most candidates knew to differentiate but some tried to substitute in $- 1.9$ for $t$ , while others struggled to differentiate the function by hand rather than using the GDC. Many candidates tried to solve the equation analytically and did not use their technology. Of those who did, many had their calculators in degree mode.

Almost all candidates who attempted part (c) thought the greatest speed was the same as the maximum of $v$ .

[N/A]