| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable resistance or force differential equation |
| Difficulty | Standard +0.8 This M4 question requires multiple sophisticated techniques: deriving acceleration from power equations, separating variables for integration with partial fractions, and using v dv/dx for distance. The algebraic manipulation is substantial and the integration non-trivial, placing it clearly above average difficulty but within reach of well-prepared Further Maths students. |
| Spec | 1.08h Integration by substitution6.02l Power and velocity: P = Fv |
A car of mass 900 kg is moving along a straight horizontal road with the engine of the car working at a constant rate of 22.5 kW. At time $t$ seconds, the speed of the car is $v$ m s$^{-1}$ ($0 \leq v \leq 30$) and the total resistance to the motion of the car has magnitude $25v$ newtons.
(a) Show that when the speed of the car is $v$ m s$^{-1}$, the acceleration of the car is $\frac{900 - v^2}{36v}$ m s$^{-2}$
(3 marks)
The time taken for the car to accelerate from 10 m s$^{-1}$ to 20 m s$^{-1}$ is $T$ seconds.
(b) Show that $T = 18\ln\frac{8}{5}$
(5 marks)
(c) Find the distance travelled by the car as it accelerates from 10 m s$^{-1}$ to 20 m s$^{-1}$
(6 marks)
---4. A car of mass 900 kg is moving along a straight horizontal road with the engine of the car working at a constant rate of 22.5 kW . At time $t$ seconds, the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 } ( 0 < v < 30 )$ and the total resistance to the motion of the car has magnitude $25 v$ newtons.
\begin{enumerate}[label=(\alph*)]
\item Show that when the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the acceleration of the car is
$$\frac { 900 - v ^ { 2 } } { 36 v } \mathrm {~m} \mathrm {~s} ^ { - 2 }$$
The time taken for the car to accelerate from $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is $T$ seconds.
\item Show that
$$T = 18 \ln \frac { 8 } { 5 }$$
\item Find the distance travelled by the car as it accelerates from $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2015 Q4 [14]}}