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Leaving Cert Applied Maths: Calculus & variable acceleration

How often Calculus & variable acceleration comes up on the Applied Maths papers, every year it was asked, and questions to try.

HL Asked on 3 of the last 3 Higher Level papers, most recently in 2025. banker

Calculus & variable acceleration, Higher Level(7 marks)

Quick ones on Calculus & variable acceleration.

(a)Given v as a function of t, how do you find displacement?
  1. Differentiate v with respect to t
  2. Integrate v with respect to t
  3. Multiply v by t
(b)Acceleration written in terms of velocity v and displacement s?
  1. a = s dv/dt
  2. a = dv/ds
  3. a = v dv/ds
(c)v = 3t² + 2 m/s. Acceleration when t = 2 s?
  1. 12 m/s²
  2. 6 m/s²
  3. 14 m/s²
Show the answers

(a) Integrate v with respect to t

(b) a = v dv/ds

(c) 12 m/s²

Your turn: Higher Level questions on Calculus & variable acceleration.

Higher Level

Asked on 3 of the last 3 Higher Level papers, most recently in 2025. banker

Every paper, year by year

YearWhere it came up
2025Q2
2024Q3
2023Q1

Links open the State Examinations Commission’s paper for that year.

More Calculus & variable acceleration questions

Calculus & variable acceleration, 3 marks

a = 6t m/s² and v = 2 m/s at t = 0. Velocity when t = 3 s?

  1. 27 m/s
  2. 18 m/s
  3. 29 m/s
Show the answer

29 m/s

Integrate: v = 3t² + c, and v = 2 at t = 0 gives c = 2. At t = 3, v = 27 + 2 = 29 m/s. Dropping c gives 27; 18 is the acceleration at t = 3.

Calculus & variable acceleration, 2 marks

Integrating ∫ t e^(2t) dt by parts. Which should you let u be?

  1. t e^(2t)
  2. t
  3. e^(2t)
Show the answer

t

Choose u to be the part that gets simpler when differentiated: t → 1. Then dv = e^(2t) dt integrates easily to ½e^(2t), and the new integral is simple.

Calculus & variable acceleration, 3 marks

v = 4 − t² m/s for t ≥ 0. Distance travelled until the particle is first at rest, 2 d.p.?

  1. 5.33 m
  2. 8.00 m
  3. 2.67 m
Show the answer

5.33 m

v = 0 at t = 2. Distance = ∫₀² (4 − t²) dt = 8 − 8/3 = 16/3 = 5.33 m. 8 forgets the t² term; 2.67 is just the 8/3 part.

Other Applied Maths topics

All of Leaving Cert Applied Maths