M2 June 2013 (R) Q3
3. A particle \(P\) moves along a straight line in such a way that at time \(t\) seconds its velocity \(v\) m s\(^{-1}\) is given by
\[v = \frac{1}{2}t^2 - 3t + 4\]Find
(a) the times when \(P\) is at rest, (4)
(b) the total distance travelled by \(P\) between \(t = 0\) and \(t = 4\). (5)
| Scheme | Marks |
|---|---|
| \(\tfrac{1}{2}t^2 - 3t + 4 = 0\) | M1 |
| \(t^2 - 6t + 8 = 0\) | |
| \((t - 2)(t - 4) = 0\) | DM1 |
| \(t = 2\) s or 4 s | A1 A1 |
| (4) |
Notes
M1 Set \(v = 0\)
DM1 Solve for \(v\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{1}{2}t^2 - 3t + 4\,\mathrm{d}t\) | M1 |
| \(= \dfrac{1}{6}t^3 - \dfrac{3}{2}t^2 + 4t\ (+C)\) | A1 |
| \(s = \displaystyle\int_0^2 \tfrac{1}{2}t^2 - 3t + 4\ \mathrm{d}t - \int_2^4 \tfrac{1}{2}t^2 - 3t + 4\ \mathrm{d}t\) | DM1 |
| \(= \left[\tfrac{1}{6}t^3 - \tfrac{3}{2}t^2 + 4t\right]_0^2 - \left[\tfrac{1}{6}t^3 - \tfrac{3}{2}t^2 + 4t\right]_2^4\) | |
| \(= \tfrac{8}{6} - 6 + 8 - (\tfrac{64}{6} - 24 + 16 - (\tfrac{8}{6} - 6 + 8))\) | A1 |
| \(= \tfrac{10}{3} - \tfrac{8}{3} + \tfrac{10}{3}\) | |
| \(= 4\) | A1 |
| (5) | |
| (9 marks) |
Notes
M1 Integration – majority of powers increasing
A1 Correct (\(+C\) not required)
DM1 Correct strategy for finging the required distance. Follow their “2”. Subtraction/swap limits/modulus signs
A1 Correct unsimplified