CAIE M1 2022 June — Question 1 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with all parameters given
DifficultyModerate -0.8 This is a straightforward kinematics problem using basic SUVAT equations and graph sketching. Part (a) requires one standard equation (v²=u²+2as then v=u+at), part (b) is routine graph sketching, and part (c) involves calculating total distance/total time. All three parts are standard textbook exercises with no problem-solving insight required, making it easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae
A car starts from rest and moves in a straight line with constant acceleration for a distance of 200 m, reaching a speed of 25 m s\(^{-1}\). The car then travels at this speed for 400 m, before decelerating uniformly to rest over a period of 5 s.
  1. Find the time for which the car is accelerating. [2]
  2. Sketch the velocity–time graph for the motion of the car, showing the key points. [2]
  3. Find the average speed of the car during its motion. [2]
Question 1:
AnswerMarks
1(a)025
200 t
AnswerMarks Guidance
2M1 uv
For use of s t or other complete method to find t
2
e.g.v2 u2 2as followed by vuat
N.B. a = 1.5625
AnswerMarks
t = 16 sA1
2
AnswerMarks Guidance
1(b)Trapezium B1
All correct through (0, 0), (16, 25), (32, 25), (37, 0)B1 FT their value of t from part (a) (t +16 + 5)
2
AnswerMarks
1(c)1

Total distance = 200400 255 [= 662.5]

AnswerMarks Guidance
2M1 1
Or trapezium 251637662.5
2
250
Or 200400 5
2
Allow their value of t from part (a) in calculating 200 and their
time from the constant speed section from part (b) in calculating
400
Allow their a from part (a) if used in calculating 200
662.5
Average speed = = 17.9 m s−1 (3 s.f.)
AnswerMarks Guidance
37A1 1325 67
Allow 17
74 74
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
--- 1(a) ---
1(a) | 025
200 t
2 | M1 | uv
For use of s t or other complete method to find t
2
e.g.v2 u2 2as followed by vuat
N.B. a = 1.5625
t = 16 s | A1
2
--- 1(b) ---
1(b) | Trapezium | B1 | Through (0, 0) and positive t-axis
All correct through (0, 0), (16, 25), (32, 25), (37, 0) | B1 | FT their value of t from part (a) (t +16 + 5)
2
--- 1(c) ---
1(c) | 1
Total distance = 200400 255 [= 662.5]
2 | M1 | 1
Or trapezium 251637662.5
2
250
Or 200400 5
2
Allow their value of t from part (a) in calculating 200 and their
time from the constant speed section from part (b) in calculating
400
Allow their a from part (a) if used in calculating 200
662.5
Average speed = = 17.9 m s−1 (3 s.f.)
37 | A1 | 1325 67
Allow 17
74 74
2
Question | Answer | Marks | Guidance
A car starts from rest and moves in a straight line with constant acceleration for a distance of 200 m, reaching a speed of 25 m s$^{-1}$. The car then travels at this speed for 400 m, before decelerating uniformly to rest over a period of 5 s.

\begin{enumerate}[label=(\alph*)]
\item Find the time for which the car is accelerating. [2]

\item Sketch the velocity–time graph for the motion of the car, showing the key points. [2]

\item Find the average speed of the car during its motion. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q1 [6]}}
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