| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Topic | Power and driving force |
| Type | Find acceleration given power |
| Difficulty | Standard +0.8 This is a multi-step mechanics problem requiring students to derive a differential equation from P=Fv and Newton's second law, then solve a separable differential equation involving integration. While the setup is standard for Further Maths mechanics, the integration and algebraic manipulation required elevates it above routine exercises. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 2.5\left[-v - 20\ln | 20-v | \right]_{10}^{15}\) — A1 |
**(i)** $\frac{80000}{v} - 4000 = 10000\frac{dv}{dt}$ — B1, B1 [2]
**(ii)** $\int dt = \int_{10}^{15} \frac{10v}{80-4v}\, dv = 2.5\int_{10}^{15} \frac{v}{20-v}\, dv$ — M1
$t = 2.5\int_{10}^{15} -1 + \frac{20}{20-v}\, dv$ — M1
$= 2.5\left[-v - 20\ln|20-v|\right]_{10}^{15}$ — A1
$= 2.5\left(\left[-15 - 20\ln 5\right] - \left[-10 - 20\ln 10\right]\right)$ — M1
$= 2.5(20\ln 2 - 5)$ or AWRT 22. — A1 [5]
**[Total: 7]**1 A lorry moves along a straight horizontal road. The engine of the lorry produces a constant power of 80 kW . The mass of the lorry is 10 tonnes and the resistance to motion is constant at 4000 N .\\
(i) Express the driving force of the lorry in terms of its velocity and hence, using Newton's second law, write down a differential equation which connects the velocity of the lorry and the time for which it has been moving.\\
(ii) Hence find the time taken, in seconds, for the lorry to accelerate from $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q1 [7]}}