CAIE M1 2022 November — Question 4 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeDistance in nth second
DifficultyStandard +0.3 This is a standard SUVAT problem requiring students to set up two equations from 'distance in nth second' and solve simultaneously for initial velocity and acceleration. While it involves algebraic manipulation and understanding of the nth second formula, it's a routine mechanics exercise with a clear method that students practice extensively in M1.
Spec3.02d Constant acceleration: SUVAT formulae

4 A particle \(P\) travels in the positive direction along a straight line with constant acceleration. \(P\) travels a distance of 52 m during the 2 nd second of its motion and a distance of 64 m during the 4th second of its motion.
  1. Find the initial speed and the acceleration of \(P\).
  2. Find the distance travelled by \(P\) during the first 10 seconds of its motion.

Question 4(a):
AnswerMarks Guidance
AnswerMark Guidance
\(52 = u(2) + 0.5a(2)^2 - \left(u(1) + 0.5a(1)^2\right)\) or \(64 = u(4) + 0.5a(4)^2 - (u(3) + 0.5a(3)^2)\)M1 Use of \(s = ut + \frac{1}{2}at^2\) or equivalent to form equation for 2nd or 4th second
M1Second equation in \(u\) and \(a\)
\(52 = u + 1.5a\) and \(64 = u + 3.5a\)A1 Two correct equations in \(u\) and \(a\)
\(12 = 2a\) leading to \(a = 6\)M1 Solves simultaneous equations to find either \(u\) or \(a\)
Initial speed \(= 43\) ms\(^{-1}\) and acceleration \(= 6\) ms\(^{-2}\)A1
Question 4(b):
AnswerMarks Guidance
AnswerMark Guidance
\(s = 43\times10 + 0.5\times6\times10^2\)M1 Use of \(s = ut + \frac{1}{2}at^2\) or equivalent
Distance \(= 730\) mA1 FT FT \(10u + 50a\) with \(u\) and \(a\) from part (a)
## Question 4(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $52 = u(2) + 0.5a(2)^2 - \left(u(1) + 0.5a(1)^2\right)$ or $64 = u(4) + 0.5a(4)^2 - (u(3) + 0.5a(3)^2)$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ or equivalent to form equation for 2nd or 4th second |
| | M1 | Second equation in $u$ and $a$ |
| $52 = u + 1.5a$ and $64 = u + 3.5a$ | A1 | Two correct equations in $u$ and $a$ |
| $12 = 2a$ leading to $a = 6$ | M1 | Solves simultaneous equations to find either $u$ or $a$ |
| Initial speed $= 43$ ms$^{-1}$ and acceleration $= 6$ ms$^{-2}$ | A1 | |

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## Question 4(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $s = 43\times10 + 0.5\times6\times10^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ or equivalent |
| Distance $= 730$ m | A1 FT | FT $10u + 50a$ with $u$ and $a$ from part (a) |

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4 A particle $P$ travels in the positive direction along a straight line with constant acceleration. $P$ travels a distance of 52 m during the 2 nd second of its motion and a distance of 64 m during the 4th second of its motion.
\begin{enumerate}[label=(\alph*)]
\item Find the initial speed and the acceleration of $P$.
\item Find the distance travelled by $P$ during the first 10 seconds of its motion.
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q4 [7]}}