OCR MEI M1 — Question 3 6 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyModerate -0.8 This is a straightforward velocity-time graph question testing basic mechanics concepts: finding acceleration from gradient, distance from area under graph, and displacement calculations. All parts use standard M1 techniques with no problem-solving insight required, making it easier than average but not trivial since it requires careful interpretation of the graph and understanding of displacement vs distance.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area
A particle travels in a straight line during the time interval \(0 \leqslant t \leqslant 12\), where \(t\) is the time in seconds. Fig. 1 is the velocity-time graph for the motion. \includegraphics{figure_3}
  1. Calculate the acceleration of the particle in the interval \(0 < t < 6\). [2]
  2. Calculate the distance travelled by the particle from \(t = 0\) to \(t = 4\). [2]
  3. When \(t = 0\) the particle is at A. Calculate how close the particle gets to A during the interval \(4 \leqslant t \leqslant 12\). [2]
Question 3:
AnswerMarks
3(i)−15 =−2.5 so −2.5 m s –2
6M1
A1Use of Δv/Δt. Condone use of v/t.
Must have - ve sign. Accept no units.2
(ii)1
×10×4=20m
AnswerMarks Guidance
2M1
A1Attempt at area or equivalent 2
(iii
AnswerMarks
)1
Area under graph is ×5×5=12.5
2
(and -ve)
AnswerMarks
closest is 20−12.5=7.5mM1
A1May be implied. Area from 4 to 9
attempted. Condone missing –ve sign. Do
not award if area
beyond 9 is used (as well).
AnswerMarks
cao2
6
mark
4(i)
AnswerMarks
= 85Area under curve
0.5´ 2´ 20+0.5´ (20+10)´ 4+0.5´ 10´ 1
AnswerMarks
mM1
resul
B1 A
expla
AnswerMarks
A1 caAttempt to find any area under curve or use const accn
ts
ny area correct (Accept 20 or 60 or 5 without
nation)
AnswerMarks
o3
(ii)
AnswerMarks
upwa20- 10
=2.5
4
AnswerMarks
rdsM1
A1
AnswerMarks
B1 AD vD/ t
accept±2.5
ccept – 2.5 downwards (allow direction specified by
AnswerMarks
diagram etc). Accept ‘opposite direction to motion’.3
(iii)v=- 2.5t+c
v = 20 when t = 2
AnswerMarks
v = - 2.5t + 25M1
M1
A1
[Allo
AnswerMarks
any vAllow their a in the form v=±at+cor v=±a(t2-) + c
cao [Allow v=202- .5 (-t2) ]
w 2/3 for different variable to t used, e.g. x. Allow
AnswerMarks
ariable name for speed]3
(iv)
AnswerMarks Guidance
Falling with negligible resistance E1 A
(v)- 1.5´ 49+ .5´ 2+72=0
- 1.5´ 36+9.5´ 6+7=10
AnswerMarks
- 1.5´ 49+9.5´ 7+ 7= 0E1
E1One of the results shown
All three shown. Be generous about the ‘show’.2
(vi)
AnswerMarks
= 81.7
ò (- 1.5t2 +9.5t+7)dt
2
[ ]
7
= - 0.5t3+4.75t2 +7t
2
æ 343 19´ 49 ö ( )
= ç - + +49÷- - 4+19+14
è 2 4 ø
AnswerMarks
25 mM1 L
A1 A
A1
A1
M1
A1
subtr
AnswerMarks
A1 caimits not required
1 for each term. Limits not required. Condone + c
Attempt to use both limits on an integrated expression
Correct substitution in their expression including
action ( may be left as an expression).
AnswerMarks
o.7
total 19
Follow through between parts of Question 5 should be allowed for the value of h (when t = 10) found in part (iii) if it is used in part (iv) or in part (v)(A).
Question 3:
--- 3(i) ---
3(i) | −15 =−2.5 so −2.5 m s –2
6 | M1
A1 | Use of Δv/Δt. Condone use of v/t.
Must have - ve sign. Accept no units. | 2
(ii) | 1
×10×4=20m
2 | M1
A1 | Attempt at area or equivalent | 2
(iii
) | 1
Area under graph is ×5×5=12.5
2
(and -ve)
closest is 20−12.5=7.5m | M1
A1 | May be implied. Area from 4 to 9
attempted. Condone missing –ve sign. Do
not award if area
beyond 9 is used (as well).
cao | 2
6
mark
4(i)
= 85 | Area under curve
0.5´ 2´ 20+0.5´ (20+10)´ 4+0.5´ 10´ 1
m | M1
resul
B1 A
expla
A1 ca | Attempt to find any area under curve or use const accn
ts
ny area correct (Accept 20 or 60 or 5 without
nation)
o | 3
(ii)
upwa | 20- 10
=2.5
4
rds | M1
A1
B1 A | D vD/ t
accept±2.5
ccept – 2.5 downwards (allow direction specified by
diagram etc). Accept ‘opposite direction to motion’. | 3
(iii) | v=- 2.5t+c
v = 20 when t = 2
v = - 2.5t + 25 | M1
M1
A1
[Allo
any v | Allow their a in the form v=±at+cor v=±a(t2-) + c
cao [Allow v=202- .5 (-t2) ]
w 2/3 for different variable to t used, e.g. x. Allow
ariable name for speed] | 3
(iv)
Falli | ng with negligible resistance | E1 A | ccept ‘zero resistance’, or ‘no resistance’ seen. | 1
(v) | - 1.5´ 49+ .5´ 2+72=0
- 1.5´ 36+9.5´ 6+7=10
- 1.5´ 49+9.5´ 7+ 7= 0 | E1
E1 | One of the results shown
All three shown. Be generous about the ‘show’. | 2
(vi)
= 81. | 7
ò (- 1.5t2 +9.5t+7)dt
2
[ ]
7
= - 0.5t3+4.75t2 +7t
2
æ 343 19´ 49 ö ( )
= ç - + +49÷- - 4+19+14
è 2 4 ø
25 m | M1 L
A1 A
A1
A1
M1
A1
subtr
A1 ca | imits not required
1 for each term. Limits not required. Condone + c
Attempt to use both limits on an integrated expression
Correct substitution in their expression including
action ( may be left as an expression).
o. | 7
total 1 | 9
Follow through between parts of Question 5 should be allowed for the value of h (when t = 10) found in part (iii) if it is used in part (iv) or in part (v)(A).
A particle travels in a straight line during the time interval $0 \leqslant t \leqslant 12$, where $t$ is the time in
seconds. Fig. 1 is the velocity-time graph for the motion.

\includegraphics{figure_3}

\begin{enumerate}[label=(\roman*)]
\item Calculate the acceleration of the particle in the interval $0 < t < 6$. [2]
\item Calculate the distance travelled by the particle from $t = 0$ to $t = 4$. [2]
\item When $t = 0$ the particle is at A. Calculate how close the particle gets to A during the interval $4 \leqslant t \leqslant 12$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q3 [6]}}
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