Question 1 Report
(a) (i) Define force and state its S.I unit.
(ii) List the two types of solid friction.
(b) A car travelling at a constant speed of \(30\ \mathrm{ms}^{-1}\) for \(20\ \mathrm{s}\) was suddenly decelerated when the driver sighted a pot-hole. It took the driver \(6\ \mathrm{s}\) to get to the pot-hole with a reduced speed of \(18\ \mathrm{ms}^{-1}\). He maintained the steady speed for another \(10\ \mathrm{s}\) to cross the pot-hole. The brakes were then applied and the car came to rest \(5\ \mathrm{s}\) later.
(i) Draw the velocity-time graph for the journey.
(ii) Calculate the deceleration during the last \(5\ \mathrm{s}\) of the journey.
(iii) Calculate the total distance covered.
(a) (i) Force is a push or pull which changes, or tends to change, the state of rest or uniform motion of a body in a straight line.
The S.I. unit of force is the newton (N).
(ii) The two types of solid friction are:
(b) (i) Velocity-time graph
The successive points plotted are t time tand tvelocity tas follows: \[(0,30),\ (20,30),\ (26,18),\ (36,18),\ (41,0).\]
(ii) Deceleration during the last 5 s
For the final stage,
\[ a=\frac{v-u}{t}=\frac{0-18}{5}=-3.6\ \text{m s}^{-2}. \]
Therefore, the deceleration is \(3.6\ \text{m s}^{-2}\).
(iii) Total distance covered
The total distance is the area under the velocity-time graph.
| Part of journey | Area under graph | Distance (m) |
|---|---|---|
| \(0\) to \(20\) s | \(30\times20\) | \(600\) |
| \(20\) to \(26\) s | \(\frac{1}{2}(30+18)\times6\) | \(144\) |
| \(26\) to \(36\) s | \(18\times10\) | \(180\) |
| \(36\) to \(41\) s | \(\frac{1}{2}\times18\times5\) | \(45\) |
\[ \text{Total distance}=600+144+180+45=\boxed{969\ \text{m}}. \]
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