Edexcel M1 2013 June — Question 5 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with algebraic unknowns
DifficultyModerate -0.3 This is a standard M1 kinematics problem involving speed-time graphs and constant acceleration equations. Parts (a)-(b) require basic graph sketching and using area under graph equals distance. Parts (c)-(d) involve setting up simultaneous equations for two vehicles meeting, which is routine M1 material. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae

A car is travelling along a straight horizontal road. The car takes 120 s to travel between two sets of traffic lights which are 2145 m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30 s until its speed is \(22 \text{ m s}^{-1}\). The car maintains this speed for \(T\) seconds. The car then moves with constant deceleration, coming to rest at the second set of traffic lights.
  1. Sketch, in the space below, a speed-time graph for the motion of the car between the two sets of traffic lights. [2]
  2. Find the value of \(T\). [3]
A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of traffic lights. The motorcycle moves from rest with constant acceleration, \(a \text{ m s}^{-2}\), and passes the car at the point \(A\) which is 990 m from the first set of traffic lights. When the motorcycle passes the car, the car is moving with speed \(22 \text{ m s}^{-1}\).
  1. Find the time it takes for the motorcycle to move from the first set of traffic lights to the point \(A\). [4]
  2. Find the value of \(a\). [2]

Part (a)
AnswerMarks Guidance
SpeedShape B1
FiguresB1
(2)
Part (b)
AnswerMarks
\(\frac{(120 + T)22}{2} = 2145\)M1 A1
\(T = 75\)A1
(3)
Part (c)
AnswerMarks
\(\frac{(t + I - 30)22}{2} = 990\)M1 A1
\(t = 60\)A1
Answer \(= 60 - 10 = 50\)A1
(4)
Part (d)
AnswerMarks
\(990 = 0.5a50^2\)M1
\(a = 0.79, 0.792, 99/125\) oeA1
(2)
[11]
Notes for Question 5:
Q5(a)
- First B1 for a trapezium starting at the origin and ending on the \(t\)-axis.
- Second B1 for the figures marked (allow missing 0 and a delineator oe for \(T\)) (allow if they have used \(T = 75\) correctly on their graph)
Q5(b)
- First M1 for producing an equation in their \(T\) only by equating the area of the trapezium to 2145, with the correct no. of terms. If using a single trapezium, we need to see evidence of using \(\frac{1}{2}\) the sum of the two parallel sides or if using triangle(s), need to see \(\frac{1}{2}\) base x height.
- Second A1 cao for a correct equation in \(T\) (This is not f.t. on their \(T\))
- Third A1 for \(T = 75\).
- N.B. Use of a single suvar equation for the whole motion of the car e.g. \(s = \frac{(u+v)}{2}\) is M0
Q5(c)
- First M1 for producing an equation in \(t\) only (they may use \((t - 30)\) oe as their variable) by equating the area of the trapezium to 990, with the correct no. of terms. If using a trapezium, we need to see evidence of using \(\frac{1}{2}\) the sum of the two parallel sides or if using triangle(s), need to see \(\frac{1}{2}\) base x height.
- First A1 for a correct equation.
- Second A1 for \(t = 60\) (Allow \(30 + 30\)).
- Third A1 for answer of 50.
- N.B. Use of a single suvar equation for the whole motion of the car e.g. \(s = \frac{(u+v)}{2}\) is M0.
- Use of the motion of the motorcycle is M0 (insufficient information).
- Use of \(v = 22\) for the motorcycle is M0.
Q5(d)
- First M1 for an equation in \(a\) only.
- First A1 for \(a = 0.79, 0.792, 99/125\) oe
- N.B. Use of \(v = 22\) for the motorcycle is M0.
## Part (a)
Speed | Shape | B1
Figures | B1
| (2)

## Part (b)
$\frac{(120 + T)22}{2} = 2145$ | M1 A1
$T = 75$ | A1
| (3)

## Part (c)
$\frac{(t + I - 30)22}{2} = 990$ | M1 A1
$t = 60$ | A1
Answer $= 60 - 10 = 50$ | A1
| (4)

## Part (d)
$990 = 0.5a50^2$ | M1
$a = 0.79, 0.792, 99/125$ oe | A1
| (2)
| [11]

**Notes for Question 5:**

**Q5(a)**
- First B1 for a trapezium starting at the origin and ending on the $t$-axis.
- Second B1 for the figures marked (allow missing 0 and a delineator oe for $T$) (allow if they have used $T = 75$ correctly on their graph)

**Q5(b)**
- First M1 for producing an equation in their $T$ only by equating the area of the trapezium to 2145, with the correct no. of terms. If using a single trapezium, we need to see evidence of using $\frac{1}{2}$ the sum of the two parallel sides or if using triangle(s), need to see $\frac{1}{2}$ base x height.
- Second A1 cao for a correct equation in $T$ (This is not f.t. on their $T$)
- Third A1 for $T = 75$.
- N.B. Use of a single suvar equation for the whole motion of the car e.g. $s = \frac{(u+v)}{2}$ is M0

**Q5(c)**
- First M1 for producing an equation in $t$ only (they may use $(t - 30)$ oe as their variable) by equating the area of the trapezium to 990, with the correct no. of terms. If using a trapezium, we need to see evidence of using $\frac{1}{2}$ the sum of the two parallel sides or if using triangle(s), need to see $\frac{1}{2}$ base x height.
- First A1 for a correct equation.
- Second A1 for $t = 60$ (Allow $30 + 30$).
- Third A1 for answer of 50.
- N.B. Use of a single suvar equation for the whole motion of the car e.g. $s = \frac{(u+v)}{2}$ is M0.
- Use of the motion of the motorcycle is M0 (insufficient information).
- Use of $v = 22$ for the motorcycle is M0.

**Q5(d)**
- First M1 for an equation in $a$ only.
- First A1 for $a = 0.79, 0.792, 99/125$ oe
- N.B. Use of $v = 22$ for the motorcycle is M0.

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A car is travelling along a straight horizontal road. The car takes 120 s to travel between two sets of traffic lights which are 2145 m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30 s until its speed is $22 \text{ m s}^{-1}$. The car maintains this speed for $T$ seconds. The car then moves with constant deceleration, coming to rest at the second set of traffic lights.

\begin{enumerate}[label=(\alph*)]
\item Sketch, in the space below, a speed-time graph for the motion of the car between the two sets of traffic lights. [2]
\item Find the value of $T$. [3]
\end{enumerate}

A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of traffic lights. The motorcycle moves from rest with constant acceleration, $a \text{ m s}^{-2}$, and passes the car at the point $A$ which is 990 m from the first set of traffic lights. When the motorcycle passes the car, the car is moving with speed $22 \text{ m s}^{-1}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the time it takes for the motorcycle to move from the first set of traffic lights to the point $A$. [4]
\item Find the value of $a$. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2013 Q5 [11]}}