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2.7 Kinematics

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Question 11

A particle moves in a straight line with constant acceleration aaa.

At time t=0t = 0t=0 the particle has velocity uuu. At time t t\,t the particle has velocity v v\,v and its displacement from its starting point is sss.

You may assume the results v=u+atv = u + atv=u+at and s=(u+v2)ts = \left(\dfrac{u + v}{2}\right)ts=(2u+v​)t.

a.

Derive the formula v2=u2+2asv^2 = u^2 + 2asv2=u2+2as.

[3]
b.

Derive the formula s=ut+12at2\displaystyle s = ut + \frac12at^2s=ut+21​at2.

[2]
Markscheme

2.7 Kinematics Questions

  1. A Level
  2. /Maths
  3. /2.7 Kinematics

102 exam-style questions on WJEC A Level Maths 2.7 Kinematics, covering 2.7.1 Kinematics, 2.7.2 Kinematics, 2.7.3 Kinematics, and 2.7.4 Kinematics. Each one has a worked solution and a mark scheme showing where the marks go.

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