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Date November 2016 Marks available 6 Reference code 16N.2.sl.TZ0.9
Level SL only Paper 2 Time zone TZ0
Command term Find and Hence Question number 9 Adapted from N/A

Question

A particle P starts from a point A and moves along a horizontal straight line. Its velocity v cms1 after t seconds is given by

v(t)={2t+2,for 0t13t+4t27,for 1t12

The following diagram shows the graph of v.

N16/5/MATME/SP2/ENG/TZ0/09

P is at rest when t=1 and t=p.

When t=q, the acceleration of P is zero.

Find the initial velocity of P.

[2]
a.

Find the value of p.

[2]
b.

(i)     Find the value of q.

(ii)     Hence, find the speed of P when t=q.

[4]
c.

(i)     Find the total distance travelled by P between t=1 and t=p.

(ii)     Hence or otherwise, find the displacement of P from A when t=p.

[6]
d.

Markscheme

valid attempt to substitute t=0 into the correct function     (M1)

eg2(0)+2

2     A1     N2

[2 marks]

a.

recognizing v=0 when P is at rest     (M1)

5.21834

p=5.22 (seconds)     A1     N2

[2 marks]

b.

(i)     recognizing that a=v     (M1)

egv=0, minimum on graph

1.95343

q=1.95     A1     N2

(ii)     valid approach to find their minimum     (M1)

egv(q), 1.75879, reference to min on graph

1.75879

speed =1.76 (cms1)     A1     N2

[4 marks]

c.

(i)     substitution of correct v(t) into distance formula,     (A1)

egp1|3t+4t27|dt, |3t+4t27dt|

4.45368

distance =4.45 (cm)     A1     N2

(ii)     displacement from t=1 to t=p (seen anywhere)     (A1)

eg4.45368, p1(3t+4t27)dt

displacement from t=0 to t=1     (A1)

eg10(2t+2)dt, 0.5×1×2, 1

valid approach to find displacement for 0tp     M1

eg10(2t+2)dt+p1(3t+4t27)dt, 10(2t+2)dt4.45

3.45368

displacement =3.45 (cm)     A1     N2

[6 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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