AS June 2025 Q2
2. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\ \text{m s}^{-1}\) in the positive \(x\)-direction, where
\[v = 3 - \sqrt{2t+1} \qquad t \geqslant 0\](a) Find the value of \(t\) when \(P\) is at instantaneous rest. (2)
The acceleration of \(P\) at time \(t\) seconds is \(a\ \text{m s}^{-2}\) in the positive \(x\)-direction.
(b) Show that \(a = \dfrac{1}{v-3}\) (3)
(c) Find the speed of \(P\) when it is decelerating at \(\dfrac{4}{3}\ \text{m s}^{-2}\) (2)
(d) Find the total distance travelled by \(P\) between \(t = 0\) and \(t = \dfrac{15}{2}\)
[Solutions relying on calculator technology are not acceptable.] (4)
[Solutions relying on calculator technology are not acceptable.] (4)
| Scheme | Marks | AO |
|---|---|---|
| \(0 = 3 - \sqrt{2t+1}\) | M1 | 2.1 |
| \(t = 4\) | A1 | 1.1b |
| (2) |
Notes
M1: Correct equation
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Differentiate \(v\) wrt \(t\) | M1 | 2.1 |
| \(a = \dfrac{-1}{\sqrt{2t+1}}\) | A1 | 1.1b |
| \(a = \dfrac{1}{v-3}\) * | A1* | 2.2a |
| (3) |
Notes
M1: Both powers decreasing by 1
A1: Correct expression
A1*: Given answer correctly obtained including “\(a =\)” seen in solution
| Scheme | Marks | AO |
|---|---|---|
| \(-\dfrac{4}{3} = \dfrac{1}{v-3}\) | M1 | 2.1 |
| \(v = 2.25\ (\text{m s}^{-1})\) | A1 | 1.1b |
| (2) |
Notes
M1: Correct equation
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Integrate \(v\) wrt \(t\) | M1 | 2.1 |
| \(3t - \dfrac{1}{3}(2t+1)^{\frac{3}{2}}\ (+C)\) | A1 | 1.1b |
| \(\left[3t - \dfrac{1}{3}(2t+1)^{\frac{3}{2}}\right]_0^4 - \left[3t - \dfrac{1}{3}(2t+1)^{\frac{3}{2}}\right]_4^{7.5}\) | M1 | 3.1a |
| \(5\dfrac{1}{6}\) (m) | A1 | 1.1b |
| (4) | ||
| (11 marks) |
Notes
M1: Both powers increasing by 1
A1: Correct expression
M1: Correct method by considering change of direction
A1: Accept 5.2 or better