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Date May 2022 Marks available 3 Reference code 22M.2.SL.TZ2.6
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 6 Adapted from N/A

Question

A particle moves in a straight line such that its velocity, vm s-1, at time t seconds is given by v=t2+1cost4, 0t3.

Determine when the particle changes its direction of motion.

[2]
a.

Find the times when the particle’s acceleration is -1.9m s-2.

[3]
b.

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

[2]
c.

Markscheme

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

t=1.57079 =π2

t=1.57 =π2 (s)             A1

 

[2 marks]

a.

recognises that at=v't             (M1)

t1=2.26277, t2=2.95736

t1=2.26, t2=2.96 (s)             A1A1

 

Note: Award M1A1A0 if the two correct answers are given with additional values outside 0t3.

 

[3 marks]

b.

speed is greatest at t=3              (A1)

a=-1.83778

a=-1.84 m s-2             A1

 

[2 marks]

c.

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.

a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules
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Topic 2—Functions » SL 2.10—Solving equations graphically and analytically
Topic 2—Functions
Topic 5 —Calculus

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