Rate of change
Ratio, proportion and rates of change
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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A distance-time graph is a straight line from $(0, 0)$ to $(5, 20)$, with time in seconds and distance in metres. Find the speed.
2 marks
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The volume of water in a jar increases from $30$ ml to $54$ ml in $8$ seconds. Find the average rate of change.
2 marks
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On a distance-time graph, a tangent at $t = 4$ rises $10$ m over a run of $5$ seconds. Estimate the speed at that time.
3 marks
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The depth of water is $10$ cm at $0$ minutes, $18$ cm at $2$ minutes, $22$ cm at $4$ minutes and $24$ cm at $6$ minutes. Find the average rates of change from $0$ to $2$ minutes and from $4$ to $6$ minutes.
3 marks
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A curve passes through $(1, 2)$ and $(3, 10)$. Find the average rate of change of $y$ with respect to $x$ between these points. A tangent at $x = 2$ has gradient $5$. What does that gradient tell you?
4 marks
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A car is $0$ m from home at $0$ s, $40$ m at $10$ s, $120$ m at $20$ s and $150$ m at $30$ s. Find the average speed for each $10$ second interval and for the whole $30$ seconds.
4 marks
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The dashed line is a tangent to the distance-time curve at $t = 6$ seconds. It passes through $(2, 5)$ and $(10, 21)$. Estimate the speed at $t = 6$.
Distance-time graph with a tangent at 6 seconds 5 marks
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The temperature of a drink is $80$°C at $0$ minutes, $60$°C at $5$ minutes, $46$°C at $10$ minutes, $36$°C at $15$ minutes and $30$°C at $20$ minutes. Find the average rates of decrease from $0$ to $5$ minutes and from $15$ to $20$ minutes. Describe how the cooling changes. Estimate the rate of decrease at $10$ minutes using the chord from $5$ to $15$ minutes.
6 marks
Mark scheme
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Answer. $4$ m/s
- The gradient is the speed.
- $\dfrac{20}{5} = 4$ m/s
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Answer. $3$ ml/s
- The increase is $24$ ml.
- $\dfrac{24}{8} = 3$ ml/s
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Answer. $2$ m/s
- The gradient of the tangent estimates the speed.
- $\dfrac{10}{5} = 2$ m/s
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Answer. $4$ cm/min and $1$ cm/min
- $\dfrac{8}{2} = 4$ cm/min
- $\dfrac{2}{2} = 1$ cm/min
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Answer. The average rate is $4$. At $x = 2$, $y$ is increasing at a rate of $5$ units of $y$ per unit of $x$.
- $\dfrac{10 - 2}{3 - 1} = 4$
- The tangent gradient is the instantaneous rate of change.
- At $x = 2$, $y$ increases by about $5$ for each increase of $1$ in $x$.
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Answer. $4$ m/s, $8$ m/s, $3$ m/s, and $5$ m/s
- $\dfrac{40}{10} = 4$ and $\dfrac{80}{10} = 8$.
- $\dfrac{30}{10} = 3$
- $\dfrac{150}{30} = 5$ m/s
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Answer. $2$ m/s
- Use the two points on the tangent.
- $\dfrac{21 - 5}{10 - 2} = 2$ m/s
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Answer. $4$°C/min, then $1.2$°C/min. The drink cools more slowly as time passes. The estimate at $10$ minutes is $2.4$°C/min.
- $\dfrac{20}{5} = 4$°C/min and $\dfrac{6}{5} = 1.2$°C/min.
- The rate of decrease gets smaller, so the drink cools more slowly.
- The chord from $5$ to $15$ minutes falls $24$°C in $10$ minutes.
- The estimated rate at $10$ minutes is $2.4$°C/min.