| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of fractional powers (standard technique), solving a factorized quadratic, and evaluating derivatives at specific points. While it involves multiple parts and careful tracking of direction changes for distance, all techniques are routine applications of calculus to kinematics with no novel problem-solving required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(v = \frac{dx}{dt}\) | M1 | Recognisable attempt to differentiate the given expression |
| \(0 = 7t^{\frac{1}{2}}\left(t^2 - 5t + 4\right)\) | DM1 | Set \(v = 0\) and solve for \(t\). Dependent on first M1 |
| \(t = 1\) and \(t = 4\) | A1 | Correct solution only |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = \ | x_1 - x_0\ | + \ |
| \(= \left\ | \frac{20}{3} - 0\right | + \left |
| \(= 56\) | A1 | Correct solution only |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(a = \frac{dv}{dt}\) | M1 | Recognisable attempt to differentiate |
| \(a = \frac{35}{2} \times 4^{\frac{3}{2}} - \frac{105}{2} \times 4^{\frac{1}{2}} + 14 \times 4^{-\frac{1}{2}}\) | M1 | Substitute \(t = 4\) in their \(a\) and simplify |
| \(= 42\) | A1 | Correct solution only |
| Total: (3) [9] |
# Question 3:
## Part 3a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $v = \frac{dx}{dt}$ | M1 | Recognisable attempt to differentiate the given expression |
| $0 = 7t^{\frac{1}{2}}\left(t^2 - 5t + 4\right)$ | DM1 | Set $v = 0$ and solve for $t$. Dependent on first M1 |
| $t = 1$ and $t = 4$ | A1 | Correct solution only |
| **Total: (3)** | | |
## Part 3b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = \|x_1 - x_0\| + \|x_4 - x_1\|$ | M1 | Correct strategy to find distance for their value(s) of $t$ in $[0,4]$. **Allow M1 if there is no change of direction in the interval** |
| $= \left\|\frac{20}{3} - 0\right| + \left|-\frac{128}{3} - \frac{20}{3}\right|$ | A1ft | Correct unsimplified expression for their distance (provided there was a change in direction in $[0,4]$). **Clearly using $x(4) + 2x(1)$ but $x(4)$ miscalculated so correct combined expression never seen. M1 only** |
| $= 56$ | A1 | Correct solution only |
| **Total: (3)** | | |
## Part 3c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $a = \frac{dv}{dt}$ | M1 | Recognisable attempt to differentiate |
| $a = \frac{35}{2} \times 4^{\frac{3}{2}} - \frac{105}{2} \times 4^{\frac{1}{2}} + 14 \times 4^{-\frac{1}{2}}$ | M1 | Substitute $t = 4$ in their $a$ and simplify |
| $= 42$ | A1 | Correct solution only |
| **Total: (3) [9]** | | |
---3. A particle $P$ moves on the $x$-axis.
At time $t = 0 , P$ is instantaneously at rest at $O$.\\
At time $t$ seconds, $t > 0$, the $x$ coordinate of $P$ is given by
$$x = 2 t ^ { \frac { 7 } { 2 } } - 14 t ^ { \frac { 5 } { 2 } } + \frac { 56 } { 3 } t ^ { \frac { 3 } { 2 } }$$
Find
\begin{enumerate}[label=(\alph*)]
\item the non-zero values of $t$ for which $P$ is at instantaneous rest
\item the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 4$
\item the acceleration of $P$ when $t = 4$\\
$\_\_\_\_$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2021 Q3 [9]}}