M1 June 2015 Q6

Notes

Allow column vectors throughout. B1 for \((-3\mathbf{i} + 4\mathbf{j})\) (m)

Notes

M1 for a clear attempt at either \(3.4\,(\mathbf{i}) = (2T - 3)\,(\mathbf{i})\) or \(-12\,(\mathbf{j}) = (4 - 5T)\,(\mathbf{j})\)

First A1 for a correct equation (either) without \(\mathbf{i}\)’s and \(\mathbf{j}\)’s

A1 for 3.2 oe

N.B. \(T = \dfrac{6.4\mathbf{i} - 16\mathbf{j}}{2\mathbf{i} - 5\mathbf{j}} = 3.2\) scores M1A1A1 BUT if RHS is not a single number, then M0. Also, if they get 3.2 and another value and don’t clearly choose 3.2 then A0

Notes

First M1 for a complete method for finding \(\mathbf{v}\)

e.g. \(\mathbf{r} = (-3\mathbf{i} + 4\mathbf{j}) + t(2\mathbf{i} - 5\mathbf{j})\) so \(\mathbf{v} = 2\mathbf{i} - 5\mathbf{j}\)

OR: \(\mathbf{v} = \dfrac{(3.4\mathbf{i} - 12\mathbf{j}) - (-3\mathbf{i} + 4\mathbf{j})}{\text{their } T}\)

OR: \(\mathbf{v} = \dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = 2\mathbf{i} - 5\mathbf{j}\)

First A1 for \(2\mathbf{i} - 5\mathbf{j}\); M1A1 can be awarded for \(2\mathbf{i} - 5\mathbf{j}\) only.

Second M1 for attempt to find magnitude of their \(\mathbf{v}\), i.e. \(\sqrt{2^2 + (-5)^2}\)

Second A1 for \(\sqrt{29}\) or 5.4 or better

OR

First M1 for attempt to find distance travelled: \(d = \sqrt{(-3 - 3.4)^2 + (4 - -12)^2}\)

First A1 if correct

Second M1 for their \(d\) / their \(T\)

Second A1 for \(\sqrt{29}\) or 5.4 or better

Alt (c)