| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Distance between two moving objects |
| Difficulty | Standard +0.3 This is a standard M4 relative velocity problem requiring vector subtraction to find relative velocity, then using position vectors and perpendicular distance formula. While it involves multiple steps (6+6 marks), the techniques are routine for mechanics students: finding relative velocity magnitude/direction, then minimizing distance using calculus or the standard closest approach formula. Slightly above average difficulty due to the two-part structure and mechanics context, but follows a well-practiced template. |
| Spec | 1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
3. Two straight horizontal roads cross at right angles at the point $X$. A girl is running, with constant speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, due north towards $X$ on one road. A car is travelling, with constant speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, due west towards $X$ on the other road.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude and direction of the velocity of the car relative to the girl, giving the direction as a bearing.\\
(6)
At noon the girl is 150 m south of $X$ and the car is 800 m east of $X$.
\item Find the shortest distance between the car and the girl during the subsequent motion.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2016 Q3 [13]}}