| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Topic | Travel graphs |
| Type | Distance from velocity function using calculus |
| Difficulty | Moderate -0.3 This is a straightforward kinematics question requiring standard calculus techniques: factorizing a cubic to find when v=0, differentiating for acceleration, and integrating for displacement. While it involves multiple parts and careful attention to signs/areas, all techniques are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
**Question 6**
**(i)** $v = t(t-2)(t-4)$ — M1 (Set $v = 0$ and attempt to solve)
$t \neq 0$ so $t = 2$ and $4$. — A1
SC: B1 for both $t = 2$ and $t = 4$ found by substitution or stated without working, and B1 if shows/explains there are no other values.
Cubic graph crossing the $t$ axis at $0$ & $2$ other places. — B1
Fully correct curve, axes and intercepts labelled and curve only between $t = 0$ and $t = 4$. — B1 **[4]**
**(ii)** $a = 3t^2 - 12t + 8$ — M1, A1 (Differentiate $v$. All terms correct. Allow if found in (i) and used here.)
$= 12 - 24 + 8 = -4\ (\text{ms}^{-2})$ — A1 **[3]** (Substitute $t = 2$. c.a.o)
**(iii)**
$x = \frac{t^4}{4} - 2t^3 + 4t^2 + c$ — M1, A1 (Integrate $v$. All terms correct; condone omission of "$+ c$". Allow definite integral as alternative.)
$x = 0$ when $t = 0$ therefore $c = 0$ — A1 (Deal with $c$ correctly or consider lower limit of definite integral.)
When $t = 2$, $x = 4 - 16 + 16 = 4$ — A1 (Indep of previous A1)
So average speed $= 4/2$ — M1 (Use formula for average speed)
$= 2\ (\text{ms}^{-1})$ — A1 **[6]** (Ft *their* $x$ when $t = 2$)
**Total: [13]**6 A particle travels along a straight line. Its velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after $t$ seconds is given by
$$v = t ^ { 3 } - 6 t ^ { 2 } + 8 t \text { for } 0 \leqslant t \leqslant 4$$
When $t = 0$ the particle is at rest at the point $P$.\\
(i) Find the times (other than $t = 0$ ) when the particle is at rest. Sketch the velocity-time graph for $0 \leqslant t \leqslant 4$.\\
(ii) Find the acceleration of the particle when $t = 2$.\\
(iii) Find an expression for the displacement of the particle from $P$ after $t$ seconds. Hence state its displacement from $P$ when $t = 2$ and find its average speed between $t = 0$ and $t = 2$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2013 Q6 [13]}}