M2 June 2007 Q8
8. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\) m s\(^{-1}\) in the direction of \(x\) increasing, where \(v\) is given by
\[v = \begin{cases} 8t - \tfrac{3}{2}t^2, & 0 \leqslant t \leqslant 4,\\[4pt] 16 - 2t, & t > 4.\end{cases}\]When \(t = 0\), \(P\) is at the origin \(O\).
Find
| Scheme | Marks |
|---|---|
| \(0 \leqslant t \leqslant 4\): \(a = 8 - 3t\) | M1 |
| \(a = 0 \Rightarrow t = 8/3\) s | DM1 |
| \(\rightarrow v = 8 \cdot \dfrac{8}{3} - \dfrac{3}{2} \cdot \left(\dfrac{8}{3}\right)^2 = \dfrac{32}{3}\) (m/s) | DM1 A1 |
| second M1 dependent on the first, and third dependent on the second. | |
| (4) |
Notes
M1 Differentiate to obtain acceleration
DM1 set acceleration. = 0 and solve for t
DM1 use their t to find the value of v
A1 32/3, 10.7 or better
OR using trial an improvement:
M1 Iterative method that goes beyond integer values
M1 Establish maximum occurs for t in an interval no bigger than \(2.5 < t < 3.5\)
M1 Establish maximum occurs for t in an interval no bigger than \(2.6 < t < 2.8\)
A1
Or M1 Find/state the coordinates of both points where the curve cuts the x axis.
DM1 Find the midpoint of these two values.
M1A1 as above.
Or M1 Convincing attempt to complete the square:
DM1 substantially correct \(8t - \dfrac{3t^2}{2} = -\dfrac{3}{2}\left(t - \dfrac{8}{3}\right)^2 + \dfrac{3}{2} \times \dfrac{64}{9}\)
DM1 Max value = constant term
A1 CSO
| Scheme | Marks |
|---|---|
| \(s = 4t^2 - t^3/2\) | M1 |
| \(t = 4\): \(\ s = 64 - 64/2 = 32\) m | M1 A1 |
| (3) |
Notes
M1 Integrate the correct expression
DM1 Substitute t = 4 to find distance (s=0 when t=0 - condone omission / ignoring of constant of integration)
A1 32(m) only
| Scheme | Marks |
|---|---|
| \(t > 4\): \(v = 0 \Rightarrow t = 8\) s | B1 |
| (1) |
Notes
B1 t = 8 (s) only
| Scheme | Marks |
|---|---|
| Either | |
| \(t > 4\) \(s = 16t - t^2\ \ (+\ C)\) | M1 |
| \(t = 4,\ s = 32 \rightarrow C = -16 \Rightarrow s = 16t - t^2 - 16\) | M1 A1 |
| \(t = 10 \rightarrow s = 44\) m | M1 A1 |
| But direction changed, so: \(t = 8,\ s = 48\) | M1 |
| Hence total dist travelled \(= 48 + 4 = 52\) m | DM1 A1 |
| (8) | |
| (16 marks) |
Notes
M1 Integrate 16-2t
M1 Use t=4, s= their value from (b) to find the value of the constant of integration. or 32 + integral with a lower limit of 4 (in which case you probably see these two marks occurring with the next two. First A1 will be for 4 correctly substituted.)
A1 \(s = 16t - t^2 - 16\) or equivalent
M1 substitute t = 10
A1 44
M1 Substitute t = 8 (their value from (c))
DM1 Calculate total distance (M mark dependent on the previous M mark.)
A1 52 (m)
OR the candidate who recognizes v = 16 – 2t as a straight line can divide the shape into two triangles:
M1 distance for t = 4 to t = candidate’s 8 = ½ \(\times\) change in time \(\times\) change in speed.
A1 8-4
A1 8-0
M1 distance for t = their 8 to t = 10 = ½ \(\times\) change in time \(\times\) change in speed.
A1 10-8
A1 0-(-4)
M1 Total distance = their (b) plus the two triangles (=32 + 16 + 4).
A1 52(m)
Or (probably accompanied by a sketch?)
| \(t=4\ \ v=8\), \(t=8\ \ v=0\), so area under line \(= \dfrac{1}{2} \times (8 - 4) \times 8\) | M1A1A1 |
| \(t=8\ \ v=0\), \(t=10\ \ v=-4\), so area above line \(= \dfrac{1}{2} \times (10 - 8) \times 4\) | M1A1A1 |
| Hence total distance \(= 32\)(from b) \(+ 16 + 4 = 52\) m. | M1A1 |
Or
| for \(t > 4\) \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = -2\), =constant \(t=4,\ v=8\); \(t=8,\ v=0\); \(t=10,\ v=-4\) | M1, A1 |
| \(s = \dfrac{u + v}{2}t = \dfrac{32}{2}t\), \(=16\) working for t = 4 to t = 8 | M1, A1 |
| \(s = \dfrac{u + v}{2}t = \dfrac{-4}{2}t\), \(=-4\) working for t = 8 to t = 10 | M1, A1 |
| total \(= 32 + 16 + 4\), \(=52\) | M1, A1 |
(Corrected from the printed mark scheme: the total is printed as 32+14+4.)