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June 2006 p4 q1
3979
A car of mass 1200 kg travels on a horizontal straight road with constant acceleration \(a \text{ m s}^{-2}\).
(i) Given that the car’s speed increases from 10 \(\text{m s}^{-1}\) to 25 \(\text{m s}^{-1}\) while travelling a distance of 525 m, find the value of \(a\).
The car’s engine exerts a constant driving force of 900 N. The resistance to motion of the car is constant and equal to \(R \text{ N}\).
(ii) Find \(R\).
Solution
(i) Use the equation of motion: \(v^2 = u^2 + 2as\).
Substitute \(v = 25 \text{ m s}^{-1}\), \(u = 10 \text{ m s}^{-1}\), and \(s = 525 \text{ m}\):
\(25^2 = 10^2 + 2a \times 525\)
\(625 = 100 + 1050a\)
\(525 = 1050a\)
\(a = \frac{525}{1050} = 0.5 \text{ m s}^{-2}\)
(ii) Apply Newton's second law: \(F = ma\).
The net force \(F\) is given by the driving force minus the resistance: \(900 - R = 1200 \times 0.5\).