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Date May 2019 Marks available 6 Reference code 19M.2.AHL.TZ2.H_6
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number H_6 Adapted from N/A

Question

A particle moves along a horizontal line such that at time t seconds, t ≥ 0, its acceleration a is given by a = 2t − 1. When t = 6 , its displacement s from a fixed origin O is 18.25 m. When t = 15, its displacement from O is 922.75 m. Find an expression for s in terms of t.

Markscheme

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

attempt to integrate a to find v              M1

v=adt=(2t1)dt

=t2t+c      A1

s=vdt=(t2t+c)dt

=t33t22+ct+d      A1

attempt at substitution of given values       (M1)

at t=6,18.25=7218+6c+d

at t=15,922.75=1125112.5+15c+d

solve simultaneously:       (M1)

c=6,d=0.25      A1

s=t33t22+6t+14

 

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.9—Kinematics problems
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Topic 5 —Calculus

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