Standard +0.3 This is a straightforward integration problem requiring students to integrate a linear acceleration function to find velocity, then determine if v=0 has solutions. The integration is simple (polynomial), and checking if the resulting quadratic equals zero is routine. Slightly above average difficulty only because it requires interpreting the discriminant result in context rather than just computing a value.
9 A particle is moving in a straight line. The acceleration \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\) of the particle at time \(t \mathrm {~s}\) is given by \(\mathrm { a } = 0.8 \mathrm { t } + 0.5\). The initial velocity of the particle is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the positive \(x\)-direction.
Determine whether the particle is ever stationary.
Use of discriminant or completing the square, showing equation has complex roots or stating that the equation has no real roots. Allow this M mark for solving their equation if it has real solutions. Ignore any reference to \(t < 0\) but do not allow stationary at \(t = 0\)
So the velocity is never zero and the particle never stationary
E1 [6]
Clear conclusion in context consistent with their working. FT their \(v\). Dependent on at least 1 method mark
OR: \(v = 0.4t^2 + 0.5t + 3\) (as above)
M1, M1, A1
As above
Uses the positivity of \(t\) to establish the positivity of \(v\)
M1
Argues that \(v\) is always positive therefore never stationary
M1, E1
Clear conclusion in context consistent with their working. FT their \(v\). Dependent on at least 1 method mark
## Question 9:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int(0.8t + 0.5)\,dt = 0.4t^2 + 0.5t + c$ | M1 | Attempt to integrate, condone omission of $+c$ |
| When $t = 0$, $v = 3$: $3 = 0.4 \times 0^2 + 0.5 \times 0 + c$ | M1 | Attempt to evaluate $c$ |
| So $v = 0.4t^2 + 0.5t + 3$ | A1 | Any form |
| Particle stationary when $v = 0$: $0.4t^2 + 0.5t + 3 = 0$ | M1 | Forming an equation using their $v = 0$ |
| discriminant $= 0.5^2 - 4 \times 0.4 \times 3 = -4.55 < 0$ | M1 | Use of discriminant or completing the square, showing equation has complex roots or stating that the equation has no real roots. Allow this M mark for solving their equation if it has real solutions. Ignore any reference to $t < 0$ but do not allow stationary at $t = 0$ |
| So the velocity is never zero and the particle never stationary | E1 [6] | Clear conclusion in context consistent with their working. FT their $v$. Dependent on at least 1 method mark |
| **OR:** $v = 0.4t^2 + 0.5t + 3$ (as above) | M1, M1, A1 | As above |
| Uses the positivity of $t$ to establish the positivity of $v$ | M1 | |
| Argues that $v$ is always positive therefore never stationary | M1, E1 | Clear conclusion in context consistent with their working. FT their $v$. Dependent on at least 1 method mark |
9 A particle is moving in a straight line. The acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of the particle at time $t \mathrm {~s}$ is given by $\mathrm { a } = 0.8 \mathrm { t } + 0.5$. The initial velocity of the particle is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$-direction.
Determine whether the particle is ever stationary.
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q9 [6]}}