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Date May 2022 Marks available 3 Reference code 22M.2.SL.TZ1.5
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number 5 Adapted from N/A

Question

A particle moves along a straight line so that its velocity, vm s-1, after t seconds is given by vt=esint+4sint for 0t6.

Find the value of t when the particle is at rest.

[2]
a.

Find the acceleration of the particle when it changes direction.

[3]
b.

Find the total distance travelled by the particle.

[2]
c.

Markscheme

recognizing at rest v=0          (M1)

t=3.34692

t=3.35 (seconds)           A1

 

Note: Award (M1)A0 for additional solutions to v=0 eg t=-0.205 or t=6.08.

 

[2 marks]

a.

recognizing particle changes direction when v=0 OR when t=3.34692          (M1)

a=-4.71439

a=-4.71ms-2           A2

 

[3 marks]

b.

distance travelled =06vdt  OR

03.34esint+4sintdt-3.346esint+4sintdt  =14.3104+6.44300          (A1)

=20.7534

=20.8 (metres)           A1

 

[2 marks]

c.

Examiners report

The majority of candidates found this question challenging but were often able to gain some of the marks in each part. However, it was not uncommon to see candidates manage either all of this question, or none of it, which unfortunately suggested that not all candidates had covered this content.

In part (a), while many recognized v=0 when the particle is at rest, a common error was to assume that t=0. It was pleasing to see the majority of those who had the correct equation, manage to progress to the correct value of t, having recognised the domain and the angle measure. Inevitably, a few candidates ignored the domain and obtained t=-0.205, or found t=0.

Part (b) was not well done. The most successful approach was to use the GDC to find the gradient of the curve at the value of t obtained in part (a). This was well communicated, concise and generally accurate, although some either rounded incorrectly, or obtained a=4.71 rather than a=-4.71. Of those that did not use the GDC, many were aware that a(t)=v'(t) and made an attempt to find an expression for v'(t). However, the majority appeared not to recognise when the particle would change direction and went on to substitute an incorrect value of t, not realising that they had obtained the required value earlier.

Those that had been successful in parts (a) and (b), particularly if they had been using their GDC, were generally able to complete part (c). However, it was disappointing that many candidates who understood that an integral was required, did not refer to the formula booklet and omitted the absolute value from the integrand.

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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