M3 June 2006 Q6
6. A particle moving in a straight line starts from rest at the point \(O\) at time \(t = 0\). At time \(t\) seconds, the velocity \(v\) m s\(^{-1}\) of the particle is given by
\[\begin{aligned} v &= 3t(t-4), &\quad 0 \leqslant t \leqslant 5,\\ v &= 75t^{-1}, &\quad 5 \leqslant t \leqslant 10. \end{aligned}\](a) Sketch a velocity-time graph for the particle for \(0 \leqslant t \leqslant 10\). (3)
(b) Find the set of values of \(t\) for which the acceleration of the particle is positive. (2)
(c) Show that the total distance travelled by the particle in the interval \(0 \leqslant t \leqslant 5\) is 39 m. (3)
(d) Find, to 3 significant figures, the value of \(t\) at which the particle returns to \(O\). (5)
| Scheme | Marks |
|---|---|
 | B1 |
| Hyperbola | B1 |
| Points | B1 |
| (3) |
Notes
The axis labels 15 and 7.5 are printed displaced in the scheme’s sketch: the maximum at \(t = 5\) is 15 and the value at \(t = 10\) is 7.5.
| Scheme | Marks |
|---|---|
| Identifying the minimum point of the parabola and 5 as the end points. | M1 |
| \(2 < t < 5\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Splitting the integral into two part, with limits 0 and 4, and 4 and 5, and evaluating both integrals. | M1 |
| \(\displaystyle\int_0^4 3t(t-4)\,\mathrm{d}t = \left[t^3 - 6t^2\right]_0^4 = -32\) and \(\displaystyle\int_4^5 3t(t-4)\,\mathrm{d}t = \left[t^3 - 6t^2\right]_4^5 = 7\) Both | A1 |
| Total distance \(= 39\) (m) * cso | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_5^{t_1} \frac{75}{t}\,\mathrm{d}t = 32 - 7\) | M1 A1 |
| \(75\left[\ln t\right]_5^{t_1} = 25\) | A1 |
| \(\ln\dfrac{t_1}{5} = \dfrac{1}{3} \Rightarrow t_1 = 5\mathrm{e}^{\frac{1}{3}}\) | M1 |
| \(\approx 6.98\) cao | A1 |
| (5) | |
| (13 marks) |