| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity function using calculus |
| Difficulty | Standard +0.3 This is a straightforward M1 kinematics question involving integration of acceleration to find velocity and displacement. Parts (i)-(iv) require standard integration techniques with given initial conditions. Part (v) asks students to repeat similar calculations and compare results—routine problem-solving with no novel insight required. Slightly easier than average due to the structured guidance and standard methods throughout. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | Integrate a to obtain v |
| Answer | Marks |
|---|---|
| Since a=0 for t >10, v=50 for t >10 | M1 |
| Answer | Marks |
|---|---|
| [4] | Attempt to integrate |
| Answer | Marks |
|---|---|
| (ii) | Continuous two part v-t graph |
| Answer | Marks |
|---|---|
| Horizontal straight line for 10≤t≤20 | B1 |
| Answer | Marks |
|---|---|
| [3] | The graph must cover t = 0 to t = 20 |
| Answer | Marks |
|---|---|
| veloc | ity |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (iii) | 1 |
| Answer | Marks |
|---|---|
| When t =10, h=667 | M1 |
| Answer | Marks |
|---|---|
| [4] | Attempt to integrate |
| Answer | Marks |
|---|---|
| (iv) | Time at constant vel =667÷50=13.3 |
| Total time t =10+13.3=23.3 | B1 |
| Answer | Marks |
|---|---|
| [2] | FT for h from part (iii) |
| Answer | Marks | Guidance |
|---|---|---|
| (v) | A | Since 500>333 |
| Answer | Marks |
|---|---|
| So there is no improvement | M1 |
| Answer | Marks |
|---|---|
| [2] | For finding the height at which the crate reaches terminal velocity, eg h = 167, or |
| Answer | Marks | Guidance |
|---|---|---|
| (v) | B | v=10t−t2 (for t≤5) |
| Answer | Marks |
|---|---|
| So better | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [3] | Integration to find v | |
| mark | comment | sub |
Question 5:
5 | (i) | Integrate a to obtain v
1
v=10t− t2 (+c)
2
t =10⇒v=100−50=50
Since a=0 for t >10, v=50 for t >10 | M1
A1
M1
A1
[4] | Attempt to integrate
Substitution of t = 10 to find v
Sound argument required for given answer. It must in some way refer to a = 0.
(ii) | Continuous two part v-t graph
80 velocity
70
60
50
40
30
20
10
time
10 20
Curve for 0≤t≤10
Horizontal straight line for 10≤t≤20 | B1
B1
B1
[3] | The graph must cover t = 0 to t = 20
B0 if no vertical scale is given
veloc | ity
time
5 | (iii) | 1
Distance fallen = ∫ 10t− t2 dt
2
1
d =5t2 − t3 +c (c=0)
6
Height = 1000 – d
1
Height = 1000−5t2 + t3
6
When t =10, h=667 | M1
A1
A1
B1
[4] | Attempt to integrate
This mark should only be given if the signs are correctly obtained.
oe
(iv) | Time at constant vel =667÷50=13.3
Total time t =10+13.3=23.3 | B1
B1
[2] | FT for h from part (iii)
FT
(v) | A | Since 500>333
The box will have reached terminal speed.
So there is no improvement | M1
A1
[2] | For finding the height at which the crate reaches terminal velocity, eg h = 167, or
equivalent relevant calculation. FT for h from part (iii) if used.
Allow either one (or both) of these two statements.
(v) | B | v=10t−t2 (for t≤5)
Terminal velocity is 25 m s-1
So better | M1
A1
A1
[3] | Integration to find v
mark | comment | subA box of emergency supplies is dropped to victims of a natural disaster from a stationary helicopter at a
height of 1000 metres. The initial velocity of the box is zero.
At time $t$ s after being dropped, the acceleration, $a\text{ m s}^{-2}$, of the box in the vertically downwards direction is
modelled by
$$a = 10 - t \text{ for } 0 \leqslant t \leqslant 10,$$
$$a = 0 \text{ for } t > 10.$$
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the velocity, $v\text{ m s}^{-1}$, of the box in the vertically downwards direction in terms of
$t$ for $0 \leqslant t \leqslant 10$.
Show that for $t > 10$, $v = 50$. [4]
\item Draw a sketch graph of $v$ against $t$ for $0 \leqslant t \leqslant 20$. [3]
\item Show that the height, $h$ m, of the box above the ground at time $t$ s is given, for $0 \leqslant t \leqslant 10$, by
$$h = 1000 - 5t^2 + \frac{1}{6}t^3.$$
Find the height of the box when $t = 10$. [4]
\item Find the value of $t$ when the box hits the ground. [2]
\item Some of the supplies in the box are damaged when the box hits the ground. So measures are considered
to reduce the speed with which the box hits the ground the next time one is dropped. Two different
proposals are made. Carry out suitable calculations and then comment on each of them.
\begin{enumerate}[label=(\alph*)]
\item The box should be dropped from a height of 500 m instead of 1000 m. [2]
\item The box should be fitted with a parachute so that its acceleration is given by
$$a = 10 - 2t \text{ for } 0 \leqslant t \leqslant 5,$$
$$a = 0 \text{ for } t > 5.$$ [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q5 [18]}}