Table 1 shows data from braking tests on the same electric car. A student uses the results to produce a speed-time graph. The mass of the car is 1200 kg. Fo...
Question 1 Report
Table 1 shows data from braking tests on the same electric car. A student uses the results to produce a speed-time graph. The mass of the car is 1200 kg. For test B, assume that the braking force is constant.
| test | initial speed / m s-1 | stopping time / s | stopping distance / m |
|---|---|---|---|
| A | 12.0 | 3.0 | 18.0 |
| B | 18.0 | 4.5 | 40.5 |
| C | 24.0 | 6.0 | 72.0 |
(a) State the SI unit of time and the SI unit of speed. [2]
(b) Calculate the magnitude of the deceleration in test B. [3]
(c) Calculate the magnitude of the resultant braking force in test B. [3]
(d) Complete the conversion: 18.0 m s-1 = ........ km h-1. [3]
(e) Explain why the acceleration of the car in test B has a negative value if forwards is defined as positive. [2]
(f) Describe the line on the speed-time graph during test B. [3]
Answer Details
(a) The SI unit of time is the second, s [1]. The SI unit of speed is metre per second, \(\mathrm{m\,s^{-1}}\) [1].
(b) For test B:
\[a=\frac{v-u}{t}=\frac{0-18.0\ \mathrm{m\,s^{-1}}}{4.5\ \mathrm{s}}=-4.0\ \mathrm{m\,s^{-2}}\]
The negative sign shows the acceleration is opposite to the defined forward direction. The magnitude of the deceleration is \(4.0\ \mathrm{m\,s^{-2}}\) [3].
(c) \[F=ma=1200\ \mathrm{kg}\times4.0\ \mathrm{m\,s^{-2}}=4800\ \mathrm{N}\]
Resultant braking force magnitude = \(4800\ \mathrm{N}\) [3].
(d) \[18.0\ \mathrm{m\,s^{-1}}\times3.6=64.8\ \mathrm{km\,h^{-1}}\]
\(18.0\ \mathrm{m\,s^{-1}}=64.8\ \mathrm{km\,h^{-1}}\) [3].
(e) The velocity is decreasing in the positive, forward direction [1]. Its acceleration acts in the opposite direction to the motion, so it has a negative value [1].
(f) The line is straight [1], sloping down from \(18.0\ \mathrm{m\,s^{-1}}\) to zero [1]. It has a constant negative gradient, showing constant deceleration [1].
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