Processing math: 100%

User interface language: English | Español

Date November 2017 Marks available 6 Reference code 17N.2.sl.TZ0.9
Level SL only Paper 2 Time zone TZ0
Command term Find 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 0t5.

When t=0, the velocity of P is 3 ms1.

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=23 (exact), 0.667, t=4     A1A1     N2

[2 marks]

a.

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

ega<0, 3t214t+80, sketch of a

correct interval     A1     N2

eg23<t<4

[2 marks]

b.

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

egva

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

t37t2+8t+c

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

eg3=037(02)+8(0)+c, c=3

v=t37t2+8t+3     A1     N6

[6 marks]

c.

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

eg0<t<23, 4<t<5, diagram

one correct substitution into distance formula     (A1)

eg230|v|, 54|v|, 423|v|, 50|v|

one correct pair     (A1)

eg3.13580 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 6 - Calculus » 6.6 » Kinematic problems involving displacement s, velocity v and acceleration a.
Show 53 related questions

View options