CAIE M1 2011 November — Question 1 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2011
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeSketching velocity-time graphs
DifficultyEasy -1.2 This is a straightforward kinematics question involving reading values from a piecewise linear velocity-time graph and applying basic formulas: acceleration = gradient, distance = area under graph. All calculations are routine with no problem-solving insight required, making it easier than average.
Spec3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

1 \includegraphics[max width=\textwidth, alt={}, center]{155bc571-80e4-4c93-859f-bb150a109211-2_675_1380_255_379} A woman walks in a straight line. The woman's velocity \(t\) seconds after passing through a fixed point \(A\) on the line is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The graph of \(v\) against \(t\) consists of 4 straight line segments (see diagram). The woman is at the point \(B\) when \(t = 60\). Find
  1. the woman's acceleration for \(0 < t < 30\) and for \(30 < t < 40\),
  2. the distance \(A B\),
  3. the total distance walked by the woman.

AnswerMarks Guidance
(i) Acceleration is 0.02 ms⁻²A1 3
Acceleration is −0.21 ms⁻²A1
M1For using the gradient property for acceleration or \(v = u + at\)
(ii) \(\frac{1}{2}(1.5 + 2.1) \times 30 + \frac{1}{2} \times 2.1 \times 10 - \frac{1}{2} \times 2.2 \times 20\)M1 For using the area property for displacement
Distance AB is 42.5 mA1 2
(iii) Total distance walked is 86.5 mB1ft 1
(i) Acceleration is 0.02 ms⁻² | A1 | 3
Acceleration is −0.21 ms⁻² | A1 |
| M1 | For using the gradient property for acceleration or $v = u + at$

(ii) $\frac{1}{2}(1.5 + 2.1) \times 30 + \frac{1}{2} \times 2.1 \times 10 - \frac{1}{2} \times 2.2 \times 20$ | M1 | For using the area property for displacement
Distance AB is 42.5 m | A1 | 2

(iii) Total distance walked is 86.5 m | B1ft | 1 | It error in '64.5' or '22.0' or both
1\\
\includegraphics[max width=\textwidth, alt={}, center]{155bc571-80e4-4c93-859f-bb150a109211-2_675_1380_255_379}

A woman walks in a straight line. The woman's velocity $t$ seconds after passing through a fixed point $A$ on the line is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The graph of $v$ against $t$ consists of 4 straight line segments (see diagram). The woman is at the point $B$ when $t = 60$. Find\\
(i) the woman's acceleration for $0 < t < 30$ and for $30 < t < 40$,\\
(ii) the distance $A B$,\\
(iii) the total distance walked by the woman.

\hfill \mbox{\textit{CAIE M1 2011 Q1 [6]}}