Unit 10

8 marks · Easy

A particle moves along a straight line so that its displacement from the starting point after t seconds is given by metres.

Triangle PQR has sides PQ = 6.4 cm, QR = 7.2 cm and angle PQR = 42°.

(a)

i. Find the velocity of the particle in terms of t.

[2 marks]

(b)

ii. Calculate the velocity when t = 4.

[2 marks]

(c)

i. Calculate the area of triangle PQR using the sides and included angle.

[3 marks]

(d)

ii. Explain why the sine rule or cosine rule is not required for this area calculation.

[1 mark]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) v = 6t + 2

(b) 26 m/s

(d) Because area = 1/2 ab sin C when you know two sides and the angle between them.

Common mistakes: Not differentiating to get velocity; Using right-angled area formula; Attempting sine rule when not needed

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Differentiates quadratic expression correctly.

A1 (a) · 1 mark · Writes v = 6t + 2 or equivalent.

M2 (b) · 1 mark · Substitutes t = 4 into velocity expression. (follow-through)

A2 (b) · 1 mark · Obtains correct value for v at t=4 (26 m/s).

M3 (c) · 1 mark · Applies formula area = 1/2 ab sin(angle) with correct values.

A3 (c) · 2 marks · Correct answer 15.3 (cm^2), 1 dp or 3 sig figs.

B1 (d) · 1 mark · Clear explanation that two sides and included angle suffice for area.

Worked solution

a) v = ds/dt = 6t + 2.

b) v at t=4: 6*4 + 2 = 24 + 2 = 26 m/s.

c) Area = 1/2 * PQ * QR * sin(PQR)

Rounding to 3 sig figs: 15.3 cm^2.

d) The area formula uses just two sides and the included angle; sine and cosine rules are for finding other sides or angles.