| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard M3 variable acceleration question requiring sketching v-t graphs, finding acceleration from differentiation, integrating to find distance (with sign consideration), and solving a transcendental equation. All techniques are routine for M3 students, though part (d) requires careful setup of displacement equations across two intervals and numerical solving. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| Parabola | B1 |
| Hyperbola | B1 |
| Points | B1 |
| (3) |
| Answer | Marks |
|---|---|
| Identifying the minimum point of the parabola and 5 as the end points. | M1 |
| \(2 < t < 5\) | A1 |
| (2) |
| Answer | Marks |
|---|---|
| Splitting the integral into two parts, with limits 0 and 4, and 4 and 5, and evaluating both integrals. | M1 |
| \(\int_0^4 3t(t-4)dt = [t^3 - 6t^2]_0^4 = -32\) and \(\int_4^5 3t(t-4)dt = [t^3 - 6t^2]_4^5 = 7\) | A1 |
| Both | A1 |
| Total distance = 39 (m) ✱ | A1 |
| (3) |
| Answer | Marks |
|---|---|
| \(\int_s^6 \frac{75}{t}dt = 32 - 7\) | M1 A1 |
| \(75[\ln t]_s^6 = 25\) | A1 |
| \(\ln\frac{t_1}{5} = \frac{1}{3} \Rightarrow t_1 = 5e^{1}\) | M1 |
| \(\approx 6.98\) | A1 |
| (5) |
| Answer | Marks |
|---|---|
| Total: [13] |
**(a)**
| Parabola | B1 |
| Hyperbola | B1 |
| Points | B1 |
| | (3) |
**(b)**
| Identifying the minimum point of the parabola and 5 as the end points. | M1 |
| $2 < t < 5$ | A1 |
| | (2) |
**(c)**
| Splitting the integral into two parts, with limits 0 and 4, and 4 and 5, and evaluating both integrals. | M1 |
| $\int_0^4 3t(t-4)dt = [t^3 - 6t^2]_0^4 = -32$ and $\int_4^5 3t(t-4)dt = [t^3 - 6t^2]_4^5 = 7$ | A1 |
| Both | A1 |
| Total distance = 39 (m) ✱ | A1 |
| | (3) |
**Note:** cso
**(d)**
| $\int_s^6 \frac{75}{t}dt = 32 - 7$ | M1 A1 |
| $75[\ln t]_s^6 = 25$ | A1 |
| $\ln\frac{t_1}{5} = \frac{1}{3} \Rightarrow t_1 = 5e^{1}$ | M1 |
| $\approx 6.98$ | A1 |
| | (5) |
**Note:** cao
| **Total:** [13] |
---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
$$v = 3t(t - 4), \quad 0 \leq t \leq 5,$$
$$v = 75t^{-1}, \quad 5 \leq t \leq 10.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch a velocity-time graph for the particle for $0 \leq t \leq 10$.
[3]
\item Find the set of values of $t$ for which the acceleration of the particle is positive.
[2]
\item Show that the total distance travelled by the particle in the interval $0 \leq t \leq 5$ is $39$ m.
[3]
\item Find, to $3$ significant figures, the value of $t$ at which the particle returns to $O$.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2006 Q6 [13]}}