Standard +0.8 This is a multi-step variable acceleration problem requiring integration twice (to find velocity then displacement), solving a quadratic to find rest times, and careful consideration of direction changes. While the calculus is standard A-level, the 7-mark allocation and need to track when the particle reverses direction elevates this above routine exercises.
A particle \(P\) moves in a straight line, passing through a point \(O\) with velocity \(4.2 \text{ ms}^{-1}\). At time \(t\) s after \(P\) passes \(O\), the acceleration, \(a \text{ ms}^{-2}\), of \(P\) is given by \(a = 0.6t - 2.7\).
Find the distance \(P\) travels between the times at which it is at instantaneous rest. [7]
coefficient in at least one term (which must be the same term).
Answer
Marks
Guidance
0.3t2 −2.7t+4.2[=0]
DM1
Set up 3TQ in t with correct constant term.
(t−2)(t−7)=0 t=2,7
A1
Both correct values of t (method not required).
Attempt to integrate v
DM1
Attempt to integrate v – increase power by 1 and a change in
coefficient in at least one term (which must be the same term) –
expression for v must be at least two terms (so may not include a
constant term) so dependent on first M mark only.
Answer
Marks
Guidance
s=0.1t3−1.35t2+4.2t [+c]
A1
For use of their positive t limits in their cubic expression for s
M1
Dependent on all previous M marks.
Using their two positive t values correctly in their three term
cubic expressions for s (cubic must contain non-zero tn terms
where n = 1,2 and 3).
Answer
Marks
Guidance
Total distance = 6.25 m
A1
For reference:
(0.123−1.3522 +4.22)
−(0.173−1.3572 +4.27)
If integration of v not explicitly shown, then this can score max
*M1 DM1 A1 then SC B1 for correct answer of 6.25 (so 4 marks
max.).
7
Question 8:
8 | v=0.3t2 −2.7t+c [c=4.2] | *M1 | Attempt to integrate a – increase power by 1 and a change in
coefficient in at least one term (which must be the same term).
0.3t2 −2.7t+4.2[=0] | DM1 | Set up 3TQ in t with correct constant term.
(t−2)(t−7)=0 t=2,7 | A1 | Both correct values of t (method not required).
Attempt to integrate v | DM1 | Attempt to integrate v – increase power by 1 and a change in
coefficient in at least one term (which must be the same term) –
expression for v must be at least two terms (so may not include a
constant term) so dependent on first M mark only.
s=0.1t3−1.35t2+4.2t [+c] | A1
For use of their positive t limits in their cubic expression for s | M1 | Dependent on all previous M marks.
Using their two positive t values correctly in their three term
cubic expressions for s (cubic must contain non-zero tn terms
where n = 1,2 and 3).
Total distance = 6.25 m | A1 | For reference:
(0.123−1.3522 +4.22)
−(0.173−1.3572 +4.27)
If integration of v not explicitly shown, then this can score max
*M1 DM1 A1 then SC B1 for correct answer of 6.25 (so 4 marks
max.).
7
A particle $P$ moves in a straight line, passing through a point $O$ with velocity $4.2 \text{ ms}^{-1}$. At time $t$ s after $P$ passes $O$, the acceleration, $a \text{ ms}^{-2}$, of $P$ is given by $a = 0.6t - 2.7$.
Find the distance $P$ travels between the times at which it is at instantaneous rest. [7]
\hfill \mbox{\textit{CAIE M1 2024 Q8 [7]}}