OCR H240/03 2018 September — Question 7 5 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2018
SessionSeptember
Marks5
TopicConstant acceleration (SUVAT)
TypeMulti-phase journey: find unknown speed or time
DifficultyModerate -0.8 This is a straightforward SUVAT problem requiring students to find three time intervals using basic kinematic equations (v² = u² + 2as for acceleration phase, then constant speed distance/time, then v = u + at for deceleration). All values are given directly with no problem-solving insight needed, making it easier than average but requiring careful bookkeeping across three phases.
Spec3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

7 \includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-07_512_1072_484_502} The diagram shows the velocity-time graph for a train travelling on a straight level track between stations \(A\) and \(B\) that are 2 km apart. The train leaves \(A\), accelerating uniformly from rest for 400 m until reaching a speed of \(32 \mathrm {~ms} ^ { - 1 }\). The train then travels at this steady speed for \(T\) seconds before decelerating uniformly at \(1.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }\), coming to rest at \(B\). Find the total time for the journey.

AnswerMarks Guidance
Time for train to decelerate is \(t_1 = \frac{32}{1.6}\)B1 \(t_1 = 20\)
Time for train to accelerate is \(t_2 = \frac{400}{\frac{1}{2} \times 32}\)B1 \(t_2 = 25\)
\(32T + \frac{1}{2}(32)t_1 = 1600 \Rightarrow T = \ldots\)M1 Setting up an equation for the area underneath the curve equal to either 2000 or 1600
A1Fully correct equation for their \(t_1\) or \(\frac{1}{2}((25 + 20 + T) + T)(32) = 2000\) and attempt to solve for \(T\)
\(T = 40\)
Time \(= t_1 + t_2 + T = 85\,\text{s}\)A1
[5 marks total]
Time for train to decelerate is $t_1 = \frac{32}{1.6}$ | B1 | $t_1 = 20$

Time for train to accelerate is $t_2 = \frac{400}{\frac{1}{2} \times 32}$ | B1 | $t_2 = 25$

$32T + \frac{1}{2}(32)t_1 = 1600 \Rightarrow T = \ldots$ | M1 | Setting up an equation for the area underneath the curve equal to either 2000 or 1600
| A1 | Fully correct equation for their $t_1$ or $\frac{1}{2}((25 + 20 + T) + T)(32) = 2000$ and attempt to solve for $T$
| | $T = 40$

Time $= t_1 + t_2 + T = 85\,\text{s}$ | A1 |

**[5 marks total]**

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\includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-07_512_1072_484_502}

The diagram shows the velocity-time graph for a train travelling on a straight level track between stations $A$ and $B$ that are 2 km apart. The train leaves $A$, accelerating uniformly from rest for 400 m until reaching a speed of $32 \mathrm {~ms} ^ { - 1 }$. The train then travels at this steady speed for $T$ seconds before decelerating uniformly at $1.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, coming to rest at $B$.

Find the total time for the journey.

\hfill \mbox{\textit{OCR H240/03 2018 Q7 [5]}}