Standard +0.3 This is a straightforward variable acceleration problem requiring two integrations (a→v→s) and solving v=0 to find when the particle stops. While it involves calculus, the setup is standard with no conceptual tricks—students simply apply the routine method of integrating acceleration, using initial conditions, then integrating velocity. The algebra is simple and the question structure is typical for M1.
A particle \(P\) moves in a straight line. It starts from rest at a point \(O\) on the line and at time \(t\) s after leaving \(O\) it has acceleration \(a \text{ m s}^{-2}\), where \(a = 6t - 18\).
Find the distance \(P\) moves before it comes to instantaneous rest. [6]
Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4
[v = 3t2 – 18t (+ C)]
*M1
[s = t3 – 9t2 (+ C)]
#M1
Attempt to integrate v
v = 3t2 – 18t
Answer
Marks
Guidance
s = t3 – 9t2
A1
Both integrals correct
v = 0, 3t2 – 18t = 0 [t = 6]
*DM1
Attempt to find t when v = 0
s = 63 – 9 × 62 – [0]
#DM1
Substitute limits correctly into s
s = 108 m
A1
Answer must be positive
6
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4 | [v = 3t2 – 18t (+ C)] | *M1 | Attempt to integrate a
[s = t3 – 9t2 (+ C)] | #M1 | Attempt to integrate v
v = 3t2 – 18t
s = t3 – 9t2 | A1 | Both integrals correct
v = 0, 3t2 – 18t = 0 [t = 6] | *DM1 | Attempt to find t when v = 0
s = 63 – 9 × 62 – [0] | #DM1 | Substitute limits correctly into s
s = 108 m | A1 | Answer must be positive
6
Question | Answer | Marks | Guidance
A particle $P$ moves in a straight line. It starts from rest at a point $O$ on the line and at time $t$ s after leaving $O$ it has acceleration $a \text{ m s}^{-2}$, where $a = 6t - 18$.
Find the distance $P$ moves before it comes to instantaneous rest. [6]
\hfill \mbox{\textit{CAIE M1 2020 Q4 [6]}}