Challenging +1.2 This question requires understanding piecewise functions, continuity conditions, finding when velocity equals zero, and integrating to find distance. While it involves multiple steps and careful algebraic manipulation across three different function forms, the techniques are all standard M1 content (integration, solving equations, substitution). The continuity condition and finding k adds modest complexity beyond routine kinematics problems, but no novel insight is required.
7 A particle moves in a straight line starting from a point \(O\) before coming to instantaneous rest at a point \(X\). At time \(t \mathrm {~s}\) after leaving \(O\), the velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) of the particle is given by
$$\begin{array} { l l }
v = 7.2 t ^ { 2 } & 0 \leqslant t \leqslant 2 , \\
v = 30.6 - 0.9 t & 2 \leqslant t \leqslant 8 , \\
v = \frac { 1600 } { t ^ { 2 } } + k t & 8 \leqslant t ,
\end{array}$$
where \(k\) is a constant. It is given that there is no instantaneous change in velocity at \(t = 8\).
Find the distance \(O X\).
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Use velocity at \(t = 8\) to set up a linear equation in \(k\) only. Allow a slip in one value or sign only
\(k = -0.2\)
A1
\(\dfrac{1600}{t^2} + (\text{their } k) \times t = 0 \Rightarrow t = \ldots\)
DM1
Attempt to find value of \(t\) when particle comes to rest using correct expression for \(v\), set equal to zero with *their* negative value of \(k\). Must find a positive value for \(t\) (for reference, \(t = 20\))
Attempt to integrate \(v\) for one of the 3 intervals
*M1
Increase power by 1 and a change in coefficient in at least one term (which must be the same term); \(s = vt\) is M0
\(s = \dfrac{7.2}{3}t^3 (+c)\)
A1
May be unsimplified (for reference, limits are from 0 to 2)
\(s = 30.6t - \dfrac{0.9}{2}t^2 (+c)\)
A1
May be unsimplified (for reference, limits are from 2 to 8)
This mark can be awarded if no integration is shown oe. e.g. \(\dfrac{1311}{5}\). Condone 262 www
9
## Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| $30.6 - 0.9 \times 8 = \dfrac{1600}{8^2} + 8k$ | ***M1** | Use velocity at $t = 8$ to set up a linear equation in $k$ only. Allow a slip in one value or sign only |
| $k = -0.2$ | **A1** | |
| $\dfrac{1600}{t^2} + (\text{their } k) \times t = 0 \Rightarrow t = \ldots$ | **DM1** | Attempt to find value of $t$ when particle comes to rest using correct expression for $v$, set equal to zero with *their* negative value of $k$. Must find a positive value for $t$ (for reference, $t = 20$) |
| Attempt to integrate $v$ for one of the 3 intervals | ***M1** | Increase power by 1 and a change in coefficient in at least one term (which must be the same term); $s = vt$ is M0 |
| $s = \dfrac{7.2}{3}t^3 (+c)$ | **A1** | May be unsimplified (for reference, limits are from 0 to 2) |
| $s = 30.6t - \dfrac{0.9}{2}t^2 (+c)$ | **A1** | May be unsimplified (for reference, limits are from 2 to 8) |
| $s = \dfrac{1600}{-1}t^{-1} + \dfrac{k}{2}t^2 (+c)$ | **A1FT** | May be unsimplified (for reference limits are from 8 to 20). Follow through *their* value of $k$ or just $k$ only |
| Either $19.2$ or $156.6$ or $\pm 86.4$ | **B1** | One correct distance found. Allow unsimplified e.g. $(216 - 59.4)$ or $\dfrac{1}{2} \times (8-2) \times (28.8 + 23.4)$ etc. |
| Distance $= 19.2 + (216 - 59.4) + (-120 - (-206.4)) = 262.2 \text{ m}$ | **B1** | This mark can be awarded if no integration is shown oe. e.g. $\dfrac{1311}{5}$. Condone 262 www |
| | **9** | |
7 A particle moves in a straight line starting from a point $O$ before coming to instantaneous rest at a point $X$. At time $t \mathrm {~s}$ after leaving $O$, the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of the particle is given by
$$\begin{array} { l l }
v = 7.2 t ^ { 2 } & 0 \leqslant t \leqslant 2 , \\
v = 30.6 - 0.9 t & 2 \leqslant t \leqslant 8 , \\
v = \frac { 1600 } { t ^ { 2 } } + k t & 8 \leqslant t ,
\end{array}$$
where $k$ is a constant. It is given that there is no instantaneous change in velocity at $t = 8$.\\
Find the distance $O X$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M1 2023 Q7 [9]}}