June 2024 Paper 1 Q12
12 In this question the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the \(x\)- and \(y\)-directions respectively.
The velocity \(\mathbf{v}\ \text{m s}^{-1}\) of a particle is given by \(\mathbf{v} = 3\mathbf{i} + (6t^2 - 5)\mathbf{j}\). The initial position of the particle is \(7\mathbf{j}\) m.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{r} = \displaystyle\int\left(3\mathbf{i} + (6t^2 - 5)\mathbf{j}\right)\mathrm{d}t\) | M1 | 1.1a |
| \(= 3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + \mathbf{c}\) | A1 | 1.1b |
| When \(t = 0, \mathbf{r}_0 = 0\mathbf{i} + 7\mathbf{j}\) | M1 | 1.1a |
| So position is \(= 3t\mathbf{i} + (2t^3 - 5t + 7)\mathbf{j}\) | A1 | 2.5 |
| [4] |
Notes
M1: Attempt to integrate velocity either as a vector or 2 separate components
A1: Condone missing constant
M1: Either as a vector constant or 2 separate components evaluated.
May be implied by correct vector answer
A1: Must be in vector form (could be column vector but must be exact vector notation eg brackets and not \(\mathbf{i}\) and \(\mathbf{j}\) as well)
Allow \(3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + 7\mathbf{j}\)
| Scheme | Marks | AO |
|---|---|---|
| Using \(x = 3t\) and \(y = 2t^3 - 5t + 7\) | M1 | 3.1a |
| We get \(y = 2\left(\dfrac{x}{3}\right)^3 - 5\left(\dfrac{x}{3}\right) + 7\) | A1 | 1.1 |
| [2] |
Notes
M1: Attempt to eliminate \(t\) from the parametric equations
A1: FT their (a)
Isw