Standard +0.3 This question requires students to find acceleration by differentiating velocity, evaluate it at t=8, then interpret the sign in context of whether speed is increasing or decreasing (noting that v is negative at t=8). While it involves multiple steps and careful sign reasoning, it's a standard mechanics application of differentiation with no novel insight required—slightly easier than average.
7 The velocity \(v \mathrm {~ms} ^ { - 1 }\) of a particle at time \(t \mathrm {~s}\) is given by
\(v = 0.5 t ( 7 - t )\).
Determine whether the speed of the particle is increasing or decreasing when \(t = 8\).
Evaluating \(a\) when \(t=8\); allow A1 for establishing \(a<0\) from correct expression without evaluating
And \(v=3.5\times 8-0.5\times 8^2=-4\)
B1
Evaluating \(v\) when \(t=8\); allow B1 for establishing \(v<0\) from correct expression without evaluating
Either "Velocity is negative and decreasing, so the speed is increasing" or "As both the velocity and acceleration are negative, the speed is increasing" or similar
A1 [4]
Argued from negative \(a\) and \(v\)
Special case:
Answer
Marks
Guidance
Answer
Marks
Guidance
When \(t=8\), \(v=3.5\times 8-0.5\times 8^2=-4\)
B1
Evaluating \(v\) when \(t=8\); method is not a full argument so max 3/4 marks
Evaluating \(v\) either side of \(t=8\)
B1
Two correct values
Statement referring to correct values
B1
Argued from correct working
## Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v=3.5t-0.5t^2\Rightarrow\frac{dv}{dt}=3.5-t$ | M1 | Attempt to differentiate to find $a$ |
| When $t=8$, $a=3.5-8=-4.5$ | A1 | Evaluating $a$ when $t=8$; allow A1 for establishing $a<0$ from correct expression without evaluating |
| And $v=3.5\times 8-0.5\times 8^2=-4$ | B1 | Evaluating $v$ when $t=8$; allow B1 for establishing $v<0$ from correct expression without evaluating |
| Either "Velocity is negative and decreasing, so the speed is increasing" or "As both the velocity and acceleration are negative, the speed is increasing" or similar | A1 [4] | Argued from negative $a$ and $v$ |
**Special case:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t=8$, $v=3.5\times 8-0.5\times 8^2=-4$ | B1 | Evaluating $v$ when $t=8$; method is not a full argument so max 3/4 marks |
| Evaluating $v$ either side of $t=8$ | B1 | Two correct values |
| Statement referring to correct values | B1 | Argued from correct working |
7 The velocity $v \mathrm {~ms} ^ { - 1 }$ of a particle at time $t \mathrm {~s}$ is given by\\
$v = 0.5 t ( 7 - t )$.
Determine whether the speed of the particle is increasing or decreasing when $t = 8$.
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q7 [4]}}