Kinematics from A-Level Physics

Classical Mechanics Level 2

A moving body undergoes uniform acceleration while traveling on a straight line between points X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and from Y to Z is 6 s.

What is the acceleration of the body?

0.37 m/s 1.1 m/s 0.56 m/s 0.49 m/s

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1 solution

Steven Chase
Nov 6, 2018

We know neither the acceleration nor the initial velocity at X X . Write two kinematic equations for the X Y XY and Y Z YZ distances.

XY: 40 = v 0 ( 12 ) + 1 2 a ( 1 2 2 ) YZ: 40 = ( v 0 + 12 a ) ( 6 ) + 1 2 a ( 6 2 ) \text{XY:} \,\, 40 = v_0 (12) + \frac{1}{2} a (12^2) \\ \text{YZ:} \,\, 40 = (v_0 + 12 a) (6) + \frac{1}{2} a (6^2)

Simplifying:

40 = 12 v 0 + 72 a 40 = 6 v 0 + 90 a 40 = 12 v_0 + 72 a \\ 40 = 6 v_0 + 90 a

Solving gives a = 10 27 ≈ 0.37 a = \frac{10}{27} \approx 0.37

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