M1 June 2017 Q6
6. A cyclist is moving along a straight horizontal road and passes a point \(A\). Five seconds later, at the instant when she is moving with speed 10 m s\(^{-1}\), she passes the point \(B\). She moves with constant acceleration from \(A\) to \(B\).
Given that \(AB = 40\) m, find
| Scheme | Marks |
|---|---|
| \(s = vt - \tfrac{1}{2}at^2\) | |
| \(40 = 10 \times 5 - \tfrac{1}{2}a5^2\) | M1 A2 |
| \(a = 0.8\) | A1 |
| (4) |
Notes
First M1 for a complete method to produce a value for \(a\). They may use two (or more equations) and solve for \(a\).(see possible equations)
A2 if all correct, A1A0 for one error
Third A1 for 0.8 (m s\(^{-2}\))
Possible equations:
\(40 = 5u + \tfrac{1}{2}a.5^2\)
\(10^2 = u^2 + 2a.40\)
\(10 = u + 5a\)
\(40 = \dfrac{(u + 10)}{2}.5\)
| Scheme | Marks |
|---|---|
| Finding \(u\) (\(= 6\)) | M1 |
| \(s = ut + \tfrac{1}{2}at^2\) (\(A\) to \(M\)) | |
| \(20 = 6t + \tfrac{1}{2}0.8t^2\) | M1 A1 |
| \(t = \dfrac{-15 \pm \sqrt{225 + 200}}{2}\) | DM1 |
| \(= 2.8\) or 2.81 or better | A1 |
| (5) | |
| (9 marks) |
Notes
First M1 for attempt to find a value for \(u\) (This may have been done in part (a) but MUST be used in (b) )
Second M1 for a complete method (may involve 2 or more suvat equations) for finding an equation in \(t\) only
First A1 for a correct equation
Third M1, dependent on previous M, for solving their equation for \(t\)
Second A1 for 2.8 (s) or better or \(\dfrac{5(2\sqrt{17} - 6)}{4}\); \(\dfrac{40}{6 + 2\sqrt{17}}\)
Alternative
| Finding \(v\) (\(= \sqrt{68}\)) | M1 |
| \(s = vt - \tfrac{1}{2}at^2\) (\(A\) to \(M\)) | |
| \(20 = \sqrt{68}t - \tfrac{1}{2}0.8t^2\) | M1 A1 |
| \(t = \dfrac{\sqrt{68} \pm \sqrt{68 - 32}}{0.8}\) | DM1 |
| \(= 2.8\) or 2.81 or better | A1 |
Alternative
| \(s = vt_1 - \tfrac{1}{2}at_1^2\) (\(M\) to \(B\)) | |
| \(20 = 10t_1 - \tfrac{1}{2}0.8t_1^2\) | M2 A1 |
| \(t_1 = \dfrac{10 \pm \sqrt{100 - 32}}{0.8}\) | DM1 |
| \(= 2.192\) | |
| \(t = 5 - t_1 = 2.8\) or 2.81 or better | A1 |