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Feb/Mar 2020 p42 q4
3355
A cyclist travels along a straight road with constant acceleration. He passes through points A, B and C. The cyclist takes 2 seconds to travel along each of the sections AB and BC and passes through B with speed 4.5 m s-1. The distance AB is \(\frac{4}{5}\) of the distance BC.
(a) Find the acceleration of the cyclist.
(b) Find AC.
Solution
(a) Use the constant acceleration equations to find expressions for \(s_{AB}\) and \(s_{BC}\):
\(s_{AB} = 2 \times 4.5 - \frac{1}{2} a \times 2^2\)
\(s_{BC} = 2 \times 4.5 + \frac{1}{2} a \times 2^2\)
Given \(s_{AB} = \frac{4}{5} s_{BC}\), substitute:
\([2 \times 4.5 - \frac{1}{2} a \times 2^2] = \frac{4}{5} [2 \times 4.5 + \frac{1}{2} a \times 2^2]\)