M3 January 2005 Q5
5. At time \(t = 0\), a particle \(P\) is at the origin \(O\), moving with speed 18 m s\(^{-1}\) along the \(x\)-axis, in the positive \(x\)-direction. At time \(t\) seconds \((t > 0)\) the acceleration of \(P\) has magnitude \(\dfrac{3}{\sqrt{(t+4)}}\) m s\(^{-2}\) and is directed towards \(O\).
(a) Show that, at time \(t\) seconds, the velocity of \(P\) is \([30 - 6\sqrt{(t+4)}]\) m s\(^{-1}\). (5)
(b) Find the distance of \(P\) from \(O\) when \(P\) comes to instantaneous rest. (7)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = -\dfrac{3}{\sqrt{t+4}}\) | M1 |
| \(v = -3\displaystyle\int (t+4)^{-\frac{1}{2}}\,\mathrm{d}t\) | |
| \(v = -6(t+4)^{\frac{1}{2}} + C\) | M1 A1 |
| \(t = 0,\ v = 18\): \(18 = -6 \times 2 + C \Rightarrow C = 30\) | M1 |
| \(v = 30 - 6\sqrt{t+4}\) * | A1 cso |
| (5) |
| Scheme | Marks |
|---|---|
| \(x = \displaystyle\int 30 - 6(t+4)^{\frac{1}{2}}\,\mathrm{d}t\) | M1 |
| \(= 30t - 4(t+4)^{\frac{3}{2}} + D\) | A1 |
| \(t = 0,\ x = 0\): \(0 = 0 - 4 \times 8 + D \Rightarrow D = 32\) | M1 |
| \(v = 0 \Rightarrow 30 - 6\sqrt{t+4} = 0 \Rightarrow t = 21\) | M1 A1 |
| When \(t = 21\), \(x = 30 \times 21 - 4 \times 5^3 + 32\) | M1 |
| \(= 162\) (m) | A1 |
| (7) | |
| (12 marks) |
Notes
(Corrected from the printed mark scheme: \(4 \times 5^3\) is printed as \(4 \times 5^2\); \((21+4)^{\frac{3}{2}} = 125\).)