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Date November 2017 Marks available 4 Reference code 17N.2.SL.TZ0.S_9
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_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 = 3 t 2 14 t + 8 , for 0 t 5 .

When t = 0 , the velocity of P is 3  m 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

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

t = 2 3  (exact),  0.667 ,   t = 4     A1A1     N2

[2 marks]

a.

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

eg a < 0 ,   3 t 2 14 t + 8 0 , sketch of a

correct interval     A1     N2

eg 2 3 < t < 4

[2 marks]

b.

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

eg v a

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

t 3 7 t 2 + 8 t + c

substituting t = 0 ,   v = 3 , (must have c )     (M1)

eg 3 = 0 3 7 ( 0 2 ) + 8 ( 0 ) + c ,   c = 3

v = t 3 7 t 2 + 8 t + 3     A1     N6

[6 marks]

c.

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

eg 0 < t < 2 3 ,   4 < t < 5 , diagram

one correct substitution into distance formula     (A1)

eg 0 2 3 | v | ,   4 5 | v | ,   2 3 4 | v | ,   0 5 | v |

one correct pair     (A1)

eg 3.13580 and 11.0833, 20.9906 and 35.2097

14.2191     A1     N2

d = 14.2  (m)

[4 marks]

d.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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