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Date November 2017 Marks available 2 Reference code 17N.2.sl.TZ0.9
Level SL only Paper 2 Time zone TZ0
Command term Write down Question number 9 Adapted from N/A

Question

Note: In this question, distance is in metres and time is in seconds.

A particle P moves in a straight line for five seconds. Its acceleration at time t is given by a=3t214t+8, for 0.

When t = 0, the velocity of P is 3{\text{ m}}\,{{\text{s}}^{ - 1}}.

Write down the values of t when a = 0.

[2]
a.

Hence or otherwise, find all possible values of t for which the velocity of P is decreasing.

[2]
b.

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

[6]
c.

Find the total distance travelled by P when its velocity is increasing.

[4]
d.

Markscheme

t = \frac{2}{3}{\text{ (exact), }}0.667,{\text{ }}t = 4     A1A1     N2

[2 marks]

a.

recognizing that v is decreasing when a is negative     (M1)

eg\,\,\,\,\,a < 0,{\text{ }}3{t^2} - 14t + 8 \leqslant 0, sketch of a

correct interval     A1     N2

eg\,\,\,\,\,\frac{2}{3} < t < 4

[2 marks]

b.

valid approach (do not accept a definite integral)     (M1)

eg\,\,\,\,\,v\int a

correct integration (accept missing c)     (A1)(A1)(A1)

{t^3} - 7{t^2} + 8t + c

substituting t = 0,{\text{ }}v = 3 , (must have c)     (M1)

eg\,\,\,\,\,3 = {0^3} - 7({0^2}) + 8(0) + c,{\text{ }}c = 3

v = {t^3} - 7{t^2} + 8t + 3     A1     N6

[6 marks]

c.

recognizing that v increases outside the interval found in part (b)     (M1)

eg\,\,\,\,\,0 < t < \frac{2}{3},{\text{ }}4 < t < 5, diagram

one correct substitution into distance formula     (A1)

eg\,\,\,\,\,\int_0^{\frac{2}{3}} {\left| v \right|,{\text{ }}\int_4^5 {\left| v \right|} ,{\text{ }}\int_{\frac{2}{3}}^4 {\left| v \right|} ,{\text{ }}\int_0^5 {\left| v \right|} }

one correct pair     (A1)

eg\,\,\,\,\,3.13580 and 11.0833, 20.9906 and 35.2097

14.2191     A1     N2

d = 14.2{\text{ (m)}}

[4 marks]

d.

Examiners report

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

Syllabus sections

Topic 6 - Calculus » 6.6 » Kinematic problems involving displacement s, velocity v and acceleration a.
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