CAIE M1 2012 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2012
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeFind acceleration from distances/times
DifficultyModerate -0.3 This is a standard two-unknown SUVAT problem requiring simultaneous equations from two intervals. While it involves algebraic manipulation, the approach is routine: apply s=ut+½at² to each interval, form two equations, and solve. The setup is straightforward with no conceptual subtlety, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae
3 A car travels along a straight road with constant acceleration \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\). It passes through points \(A , B\) and \(C\); the time taken from \(A\) to \(B\) and from \(B\) to \(C\) is 5 s in each case. The speed of the car at \(A\) is \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the distances \(A B\) and \(B C\) are 55 m and 65 m respectively. Find the values of \(a\) and \(u\). [6]
Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
M1For using \(s = ut + \frac{1}{2}at^2\) for \(AB\) or \(AC\)
\(55 = 5u + 12.5a\)A1
\((55 + 65) = 10u + 50a\) or \(65 = 5v_B + 12.5a\) and \(v_B = u + 5a\)A1
M1For solving for \(a\) or \(u\)
\(a = 0.4\) (or \(u = 10\))A1
\(u = 10\) (or \(a = 0.4\))A1ft [6]
Alternative:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(v_B = (55 + 65) \div (5 + 5)\)M1 For calculating speed at \(B\) as mean speed for motion from \(A\) to \(C\)
\(v_B = 12\ \text{ms}^{-1}\)A1
\(55 \div 5 = 11\ \text{ms}^{-1}\)B1 Speed at \(X\), where \(X\) is point where car passes 2.5 s after passing through \(A\)
\([a = (12 - 11) \div 2.5]\)M1 For using \(a = (v_B - v_X) \div 2.5\)
\(a = 0.4\)A1
\(u = v_X - a \times 2.5 = 11 - 0.4 \times 2.5 = 10\)B1 
# Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $s = ut + \frac{1}{2}at^2$ for $AB$ or $AC$ |
| $55 = 5u + 12.5a$ | A1 | |
| $(55 + 65) = 10u + 50a$ or $65 = 5v_B + 12.5a$ and $v_B = u + 5a$ | A1 | |
| | M1 | For solving for $a$ or $u$ |
| $a = 0.4$ (or $u = 10$) | A1 | |
| $u = 10$ (or $a = 0.4$) | A1ft | **[6]** |

### Alternative:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_B = (55 + 65) \div (5 + 5)$ | M1 | For calculating speed at $B$ as mean speed for motion from $A$ to $C$ |
| $v_B = 12\ \text{ms}^{-1}$ | A1 | |
| $55 \div 5 = 11\ \text{ms}^{-1}$ | B1 | Speed at $X$, where $X$ is point where car passes 2.5 s after passing through $A$ |
| $[a = (12 - 11) \div 2.5]$ | M1 | For using $a = (v_B - v_X) \div 2.5$ |
| $a = 0.4$ | A1 | |
| $u = v_X - a \times 2.5 = 11 - 0.4 \times 2.5 = 10$ | B1 | |

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3 A car travels along a straight road with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$. It passes through points $A , B$ and $C$; the time taken from $A$ to $B$ and from $B$ to $C$ is 5 s in each case. The speed of the car at $A$ is $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the distances $A B$ and $B C$ are 55 m and 65 m respectively. Find the values of $a$ and $u$. [6]

\hfill \mbox{\textit{CAIE M1 2012 Q3 [6]}}
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