OCR PURE — Question 9 4 marks

Exam BoardOCR
ModulePURE
Marks4
PaperDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeMulti-phase journey: find unknown speed or time
DifficultyModerate -0.8 This is a straightforward SUVAT problem requiring basic application of kinematic equations across three stages of motion. Part (a) uses v=u+at directly, part (b) is a standard sketch, and part (c) involves finding distance in each stage and solving a simple equation. All steps are routine with no problem-solving insight required.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

9 A cyclist travels along a straight horizontal road between house \(A\) and house \(B\). The cyclist starts from rest at \(A\) and moves with constant acceleration for 20 seconds, reaching a velocity of \(15 \mathrm {~ms} ^ { - 1 }\). The cyclist then moves at this constant velocity before decelerating at \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\), coming to rest at \(B\).
  1. Find the time, in seconds, for which the cyclist is decelerating.
  2. Sketch a velocity-time graph for the motion of the cyclist between \(A\) and \(B\). [Your sketch need not be drawn to scale; numerical values need not be shown.] The total distance between \(A\) and \(B\) is 1950 m .
  3. Find the time, in seconds, for which the cyclist is moving at constant velocity.

Question 9:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
\(50\) (s)B1 3.4
Total: [1]
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
Correct \((t,v)\) graph: trapezium shape rising then flat then fallingB1 1.1 - Correct \((t,v)\) graph; no values on axes required
Total: [1]
Part (c):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{2}(15)(20) + 15T + \frac{1}{2}(15)(\text{'50'}) = 1950\) or \(\frac{1}{2}(20+T+\text{'50'}+T)\times 15 = 1950\)M1 3.4 - "Correct" equation for finding required time \(T\) using their (a); may not be earned until e.g. \(\frac{\text{"1425"}}{15}\) seen
\(95\) (s)A1 1.1
Total: [2]
# Question 9:

## Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $50$ (s) | B1 | 3.4 |

**Total: [1]**

## Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct $(t,v)$ graph: trapezium shape rising then flat then falling | B1 | 1.1 - Correct $(t,v)$ graph; no values on axes required |

**Total: [1]**

## Part (c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}(15)(20) + 15T + \frac{1}{2}(15)(\text{'50'}) = 1950$ or $\frac{1}{2}(20+T+\text{'50'}+T)\times 15 = 1950$ | M1 | 3.4 - "Correct" equation for finding required time $T$ using their (a); may not be earned until e.g. $\frac{\text{"1425"}}{15}$ seen |
| $95$ (s) | A1 | 1.1 |

**Total: [2]**

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9 A cyclist travels along a straight horizontal road between house $A$ and house $B$.

The cyclist starts from rest at $A$ and moves with constant acceleration for 20 seconds, reaching a velocity of $15 \mathrm {~ms} ^ { - 1 }$. The cyclist then moves at this constant velocity before decelerating at $0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, coming to rest at $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the time, in seconds, for which the cyclist is decelerating.
\item Sketch a velocity-time graph for the motion of the cyclist between $A$ and $B$. [Your sketch need not be drawn to scale; numerical values need not be shown.]

The total distance between $A$ and $B$ is 1950 m .
\item Find the time, in seconds, for which the cyclist is moving at constant velocity.
\end{enumerate}

\hfill \mbox{\textit{OCR PURE  Q9 [4]}}