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Further Kinematics

Further Kinematics

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Question 320

A particle moves in a straight line with an initial velocity of 5 m s−15 \text{ m s}^{-1}5 m s−1.

The acceleration a m s−2a \text{ m s}^{-2}a m s−2 of the particle at time ttt seconds is given by

a=6kt2−4kt+2 a = 6kt^2 - 4kt + 2 a=6kt2−4kt+2

where kkk is a constant.

When t=2t = 2t=2, the velocity of the particle is 13 m s−113 \text{ m s}^{-1}13 m s−1.

Show that k=12k = \frac{1}{2}k=21​.

[5]
Markscheme

Further Kinematics Questions

  1. A Level
  2. /Maths
  3. /Further Kinematics

363 exam-style questions on Edexcel A Level Maths Further Kinematics, covering 8.1 Vectors in Kinematics, 8.2 Vector Methods with Projectiles, 8.3 Variable Acceleration in One Dimension, 8.4 Differentiating Vectors, and 8.5 Integrating Vectors. Each one has a worked solution and a mark scheme showing where the marks go.

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