CAIE M1 2024 March — Question 5 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionMarch
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeTotal distance with direction changes
DifficultyStandard +0.3 This is a straightforward variable acceleration question requiring integration of a polynomial velocity function and understanding of when to use absolute values for distance. Part (a) is direct integration over a positive velocity interval. Part (b) requires differentiating to find acceleration, solving a quadratic, then integrating with attention to sign changes—all standard M1 techniques with no novel insight required. Slightly easier than average due to clear guidance about velocity signs.
Spec1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration
A particle moves in a straight line starting from a point \(O\). The velocity \(v\) m s\(^{-1}\) of the particle \(t\) s after leaving \(O\) is given by $$v = t^3 - \frac{9}{2}t^2 + 1 \text{ for } 0 \leqslant t \leqslant 4.$$ You may assume that the velocity of the particle is positive for \(t < \frac{1}{2}\), is zero at \(t = \frac{1}{2}\) and is negative for \(t > \frac{1}{2}\).
  1. Find the distance travelled between \(t = 0\) and \(t = \frac{1}{2}\). [4]
  2. Find the positive value of \(t\) at which the acceleration is zero. Hence find the total distance travelled between \(t = 0\) and this instant. [4]
Question 5:
AnswerMarks Guidance
5(a)For attempt at integration *M1
least one term (which must be the same term).
s = vt is M0.
1 9 1 3
(s=) t3+1− t2+1+t +c = t4 − t3+t +c 
AnswerMarks Guidance
(3+1) 2(2+1) 4 2A1 May be unsimplified.
 4 3 
1 1 31 1
Distance =     −   +     −(0) 
4 2 22 2
AnswerMarks Guidance
 DM1 1 1
Use limits 0 and correctly, or substitute t = .
2 2
M0 if including other regions.
21
= m
AnswerMarks Guidance
64A1 Oe, e.g. 0.328125. Condone 0.328.
Special Case if no integration seen. Maximum 1/4
21
m
AnswerMarks Guidance
64B1 Oe, e.g. 0.328125. Condone 0.328.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
5(b)Attempt to differentiate (a=)3t2 −9t
 *M1 Decrease power by 1 and a change in coefficient in at
least one term (which must be the same term).
v
a= is M0.
t
AnswerMarks Guidance
Solve a=0 to get t=3A1 Ignore t=0 if not rejected.
1
Distance from t = to t =3 =
2
  4 3 
1 3  1 1 31 1
  34 − 33 +3−   −   + =
4 2   4 2 22 2 
  
1 3  21 37 1125 
  34 − 33+3 −  = 17 = = 17.578125 
AnswerMarks Guidance
4 2  64 64 64 DM1 Must be using an expression for s from integration.
1
Allow missing minus sign at start; use limits and
2
1
their 3 correctly, where their 34.
2
21
Or use limit 3 and find difference from their
64
from part (a).
May see
 4 3 
1 1 31 1 1 3 
2   −   + − 34 − 33+3
 
4 2 22 2 4 2 
 
 21 37 29
So total = +17 =17 m
 
AnswerMarks Guidance
 64 64 32A1FT 573
Oe, e.g. , 17.90625. Condone 17.9.
32
FT their positive integration value in part (a),
37 69
e.g. 17 +their (a) or +2 their (a).
64 4
AnswerMarks Guidance
QuestionAnswer Marks
5(b)Special Case if no integration seen. Maximum M1A1B1 for 3 marks
Attempt to differentiate (a=)3t2 −9t
AnswerMarks Guidance
 M1 Decrease power by 1 and a change in coefficient in at
least one term (which must be the same term).
v
a= is M0.
t
AnswerMarks
Solve a=0 to get t=3A1
29
17 m
AnswerMarks Guidance
32B1 573
Oe, e.g. , 17.90625. Condone 17.9.
32
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | For attempt at integration | *M1 | Increase power by 1 and a change in coefficient in at
least one term (which must be the same term).
s = vt is M0.
1 9 1 3
(s=) t3+1− t2+1+t +c = t4 − t3+t +c 
(3+1) 2(2+1) 4 2 | A1 | May be unsimplified.
 4 3 
1 1 31 1
Distance =     −   +     −(0) 
4 2 22 2
  | DM1 | 1 1
Use limits 0 and correctly, or substitute t = .
2 2
M0 if including other regions.
21
= m
64 | A1 | Oe, e.g. 0.328125. Condone 0.328.
Special Case if no integration seen. Maximum 1/4
21
m
64 | B1 | Oe, e.g. 0.328125. Condone 0.328.
4
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | Attempt to differentiate (a=)3t2 −9t
  | *M1 | Decrease power by 1 and a change in coefficient in at
least one term (which must be the same term).
v
a= is M0.
t
Solve a=0 to get t=3 | A1 | Ignore t=0 if not rejected.
1
Distance from t = to t =3 =
2
  4 3 
1 3  1 1 31 1
  34 − 33 +3−   −   + =
4 2   4 2 22 2 
  
1 3  21 37 1125 
  34 − 33+3 −  = 17 = = 17.578125 
4 2  64 64 64  | DM1 | Must be using an expression for s from integration.
1
Allow missing minus sign at start; use limits and
2
1
their 3 correctly, where their 34.
2
21
Or use limit 3 and find difference from their
64
from part (a).
May see
 4 3 
1 1 31 1 1 3 
2   −   + − 34 − 33+3
 
4 2 22 2 4 2 
 
 21 37 29
So total = +17 =17 m
 
 64 64 32 | A1FT | 573
Oe, e.g. , 17.90625. Condone 17.9.
32
FT their positive integration value in part (a),
37 69
e.g. 17 +their (a) or +2 their (a).
64 4
Question | Answer | Marks | Guidance
5(b) | Special Case if no integration seen. Maximum M1A1B1 for 3 marks
Attempt to differentiate (a=)3t2 −9t
  | M1 | Decrease power by 1 and a change in coefficient in at
least one term (which must be the same term).
v
a= is M0.
t
Solve a=0 to get t=3 | A1
29
17 m
32 | B1 | 573
Oe, e.g. , 17.90625. Condone 17.9.
32
4
Question | Answer | Marks | Guidance
A particle moves in a straight line starting from a point $O$. The velocity $v$ m s$^{-1}$ of the particle $t$ s after leaving $O$ is given by

$$v = t^3 - \frac{9}{2}t^2 + 1 \text{ for } 0 \leqslant t \leqslant 4.$$

You may assume that the velocity of the particle is positive for $t < \frac{1}{2}$, is zero at $t = \frac{1}{2}$ and is negative for $t > \frac{1}{2}$.

\begin{enumerate}[label=(\alph*)]
\item Find the distance travelled between $t = 0$ and $t = \frac{1}{2}$. [4]
\item Find the positive value of $t$ at which the acceleration is zero. Hence find the total distance travelled between $t = 0$ and this instant. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q5 [8]}}
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