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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 = 2 t − 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 = a d t = ( 2 t 1 ) d t

= t 2 t + c       A1

s = v d t = ( t 2 t + c ) d t

= t 3 3 t 2 2 + c t + d       A1

attempt at substitution of given values       (M1)

at  t = 6 , 18.25 = 72 18 + 6 c + d

at  t = 15 , 922.75 = 1125 112.5 + 15 c + d

solve simultaneously:       (M1)

c = 6 , d = 0.25       A1

s = t 3 3 t 2 2 + 6 t + 1 4

 

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.9—Kinematics problems
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