OCR M1 2013 June — Question 4 10 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyStandard +0.3 This is a standard M1 velocity-time graph question requiring straightforward application of SUVAT equations and area-under-graph calculations. While it has three parts and multiple stages of motion, each step follows directly from the previous one using routine mechanics techniques (v = u + at, area = distance). No novel problem-solving insight is required, making it slightly easier than average.
Spec3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

4 \includegraphics[max width=\textwidth, alt={}, center]{b7f05d10-9d3c-4098-846d-ca6511c75c5d-3_298_540_262_735} The diagram shows the \(( t , v )\) graph of a car moving along a straight road, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of the car at time \(t \mathrm {~s}\) after it passes through the point \(A\). The car passes through \(A\) with velocity \(18 \mathrm {~ms} ^ { - 1 }\), and moves with constant acceleration \(2.4 \mathrm {~ms} ^ { - 2 }\) until \(t = 5\). The car subsequently moves with constant velocity until it is 300 m from \(A\). When the car is more than 300 m from \(A\), it has constant deceleration \(6 \mathrm {~ms} ^ { - 2 }\), until it comes to rest.
  1. Find the greatest speed of the car.
  2. Calculate the value of \(t\) for the instant when the car begins to decelerate.
  3. Calculate the distance from \(A\) of the car when it is at rest.

Question 4(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Distance \(= 18t\)B1 Lower portion of area
Distance \(= (t + [t-5]) \times (30-18)/2\)B1 Upper portion of area
\(18t + (t + [t-5]) \times (30-18)/2 = 300\)M1A1 \(30t - 30 = 300\)
\(t = 11\)A1
## Question 4(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= 18t$ | B1 | Lower portion of area |
| Distance $= (t + [t-5]) \times (30-18)/2$ | B1 | Upper portion of area |
| $18t + (t + [t-5]) \times (30-18)/2 = 300$ | M1A1 | $30t - 30 = 300$ |
| $t = 11$ | A1 | |

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4\\
\includegraphics[max width=\textwidth, alt={}, center]{b7f05d10-9d3c-4098-846d-ca6511c75c5d-3_298_540_262_735}

The diagram shows the $( t , v )$ graph of a car moving along a straight road, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of the car at time $t \mathrm {~s}$ after it passes through the point $A$. The car passes through $A$ with velocity $18 \mathrm {~ms} ^ { - 1 }$, and moves with constant acceleration $2.4 \mathrm {~ms} ^ { - 2 }$ until $t = 5$. The car subsequently moves with constant velocity until it is 300 m from $A$. When the car is more than 300 m from $A$, it has constant deceleration $6 \mathrm {~ms} ^ { - 2 }$, until it comes to rest.\\
(i) Find the greatest speed of the car.\\
(ii) Calculate the value of $t$ for the instant when the car begins to decelerate.\\
(iii) Calculate the distance from $A$ of the car when it is at rest.

\hfill \mbox{\textit{OCR M1 2013 Q4 [10]}}