C2 January 2006 Q6
6. The speed, \(v\) m s–1, of a train at time \(t\) seconds is given by\[v = \sqrt{(1.2^t - 1)}, \quad 0 \leqslant t \leqslant 30.\]
The following table shows the speed of the train at 5 second intervals.
| \(t\) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| \(v\) | 0 | 1.22 | 2.28 | 6.11 |
(a) Complete the table, giving the values of \(v\) to 2 decimal places. (3)
The distance, \(s\) metres, travelled by the train in 30 seconds is given by\[s = \int_0^{30} \sqrt{(1.2^t - 1)}\,\mathrm{d}t.\]
(b) Use the trapezium rule, with all the values from your table, to estimate the value of \(s\). (3)
| \(t\) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| \(v\) | 0 | 1.22 | 2.28 | 3.80 | 6.11 | 9.72 | 15.37 |
| Scheme | Marks |
|---|---|
| \(t = 15 \quad 25 \quad 30\) \(v = \underline{3.80} \quad \underline{9.72} \quad \underline{15.37}\) | B1 B1 B1 |
| (3) |
Notes
S.C. Penalise AWRT these values once at first offence, thus the following marks could be AWRT 2 dp (Max 2/3)
| Scheme | Marks |
|---|---|
| \(S \approx \tfrac{1}{2} \times 5; [0 + 15.37 + 2(1.22 + 2.28 + 3.80 + 6.11 + 9.72)]\) | B1 [M1] |
| \(= \tfrac{5}{2}[61.63] = 154.075 =\) AWRT \(\underline{154}\) | A1 |
| (3) | |
| (6 marks) |