CAIE M1 2019 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2019
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyModerate -0.8 This is a straightforward velocity-time graph question requiring basic kinematics: part (i) uses acceleration = change in velocity / time to find V, and part (ii) uses area under graph = distance to find U. Both parts involve simple algebraic manipulation with no conceptual challenges beyond standard GCSE/AS-level mechanics.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

\includegraphics{figure_2} The diagram shows a velocity-time graph which models the motion of a tractor. The graph consists of four straight line segments. The tractor passes a point \(O\) at time \(t = 0\) with speed \(U\) m s\(^{-1}\). The tractor accelerates to a speed of \(V\) m s\(^{-1}\) over a period of 5 s, and then travels at this speed for a further 25 s. The tractor then accelerates to a speed of 12 m s\(^{-1}\) over a period of 5 s. The tractor then decelerates to rest over a period of 15 s.
  1. Given that the acceleration of the tractor between \(t = 30\) and \(t = 35\) is 0.8 m s\(^{-2}\), find the value of \(V\). [2]
  2. Given also that the total distance covered by the tractor in the 50 seconds of motion is 375 m, find the value of \(U\). [3]

Question 2:

AnswerMarks
2(i)( )
12−V
=0.8 or 12 = V + 0.8 × 5
( )
AnswerMarks Guidance
35−30M1 Use gradient of graph or constant acceleration formulae to set up
an equation in V
AnswerMarks
V=8A1
2

AnswerMarks
2(ii) 25×8+5×10+15×6+ 1 ×( U +8 )×5=375 
 
AnswerMarks Guidance
 2 M1 Attempt to find total distance travelled by the tractor in 50s to
set up an equation for U using EITHER areas
OR suvat equations OR a combination of areas and suvat
In either case total distance must be attempted
AnswerMarks
A1FTCorrect equation FT on their V from (i)
U=6A1
3
AnswerMarks Guidance
QuestionAnswer Mark
Question 2:
--- 2(i) ---
2(i) | ( )
12−V
=0.8 or 12 = V + 0.8 × 5
( )
35−30 | M1 | Use gradient of graph or constant acceleration formulae to set up
an equation in V
V=8 | A1
2
--- 2(ii) ---
2(ii) |  25×8+5×10+15×6+ 1 ×( U +8 )×5=375 
 
 2  | M1 | Attempt to find total distance travelled by the tractor in 50s to
set up an equation for U using EITHER areas
OR suvat equations OR a combination of areas and suvat
In either case total distance must be attempted
A1FT | Correct equation FT on their V from (i)
U=6 | A1
3
Question | Answer | Mark | Guidance
\includegraphics{figure_2}

The diagram shows a velocity-time graph which models the motion of a tractor. The graph consists of four straight line segments. The tractor passes a point $O$ at time $t = 0$ with speed $U$ m s$^{-1}$. The tractor accelerates to a speed of $V$ m s$^{-1}$ over a period of 5 s, and then travels at this speed for a further 25 s. The tractor then accelerates to a speed of 12 m s$^{-1}$ over a period of 5 s. The tractor then decelerates to rest over a period of 15 s.

\begin{enumerate}[label=(\roman*)]
\item Given that the acceleration of the tractor between $t = 30$ and $t = 35$ is 0.8 m s$^{-2}$, find the value of $V$. [2]

\item Given also that the total distance covered by the tractor in the 50 seconds of motion is 375 m, find the value of $U$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2019 Q2 [5]}}