June 2025 Paper 1 Q6
6 A car is travelling along a straight horizontal road. The car’s velocity \(v\,\mathrm{m\,s^{-1}}\) at time \(t\) s is shown in the velocity-time graph below. The points (0, 3) and (5, 3) are joined with a line segment, and the points (5, 3) and (15, \(-2\)) are joined with another line segment.
The car is then attached to a caravan by means of a light inextensible horizontal tow bar and continues travelling along the road. You are given the following information.
- The car and caravan accelerate at \(1.5\,\mathrm{m\,s^{-2}}\).
- The mass of the car is 1400 kg and the mass of the caravan is 900 kg.
- The driving force acting on the car is \(D\) N and the tension in the tow bar is \(T\) N.
- The resistances to motion acting on the car and caravan are 400 N and 450 N respectively.
| Scheme | Marks | AO |
|---|---|---|
| \(v=0\) when \(t=11\) s | B1 | 3.1b |
| Distance \(= 5\times 3 + \frac{1}{2}\times 6\times 3 + \frac{1}{2}\times 4\times 2\) | M1 | 1.1 |
| \([=15+9+4] = 28\) [m] | A1 | 1.1 |
| [3] |
Notes
B1: soi. May be implied by \(t=6\) used in suvat
M1: Attempt to find the area between the graph and the \(x\)-axis
Allow \(\pm 2\) used for height of the triangle below the axis
soi
Trapezium plus triangle gives \(\frac{1}{2}\times 3(11+5)\) instead of the first two terms
A1: cao
Alternative method for M1A1
| Scheme | Marks |
|---|---|
| Distance \(= 5\times 3 + \left(6\times 3 + \frac{1}{2}\left(-\frac{1}{2}\right)\times 6^2\right) + \left|\frac{1}{2}\left(-\frac{1}{2}\right)\times 4^2\right|\) | M1 |
| \([=15+9+4] = 28\) [m] | A1 |
M1: Using suvat equation(s) with \(a = -\frac{5}{10}\). Award M1 even if \(t=10\) used
| Scheme | Marks | AO |
|---|---|---|
| Car \(D - T - 400 = 1400\times 1.5\) | B1 | 1.1a |
| Caravan \(T - 450 = 900\times 1.5\) | B1 | 1.1a |
| [2] |
Notes
B1: No extra forces
Allow \(a\) used for 1.5 Any form
RHS may be 2100
B1: No extra forces
Allow \(a\) used for 1.5 Any form
RHS may be 1350
| Scheme | Marks | AO |
|---|---|---|
| Caravan \(T-450 = 900\times 1.5\) gives \(T = 1800\) | B1 | 1.1 |
| Add the equations \(D - 850 = 2300\times 1.5\) | M1 | 1.1 |
| \(D = 4300\) | A1 | 1.1 |
| [3] |
Notes
B1: cao
M1: Solving their equations leading to a value for \(D\) soi
May see use of \(T = 1800\) directly.
Allow for whole system equation soi
A1: cao