ExpertMedium2 marksStructured
AQA GCSE · Question 21.1 · Algebra
The graph represents the velocity of a ball as it rolls along the ground.
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Work out an estimate for the acceleration of the ball, in m/s², after 2 seconds. You must show your working.
The graph represents the velocity of a ball as it rolls along the ground.
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Work out an estimate for the acceleration of the ball, in m/s², after 2 seconds. You must show your working.
How to approach this question
1. Understand that acceleration is the gradient of a velocity-time graph.
2. Since the graph is a curve at t=2 seconds, you cannot find the exact gradient. You must estimate it by drawing a tangent.
3. Carefully draw a straight line that just touches the curve at the point where t=2. Make the line reasonably long so you can pick two easy-to-read points on it.
4. Choose two points on your tangent line (not on the curve itself) and read their coordinates.
5. Calculate the gradient using the formula: (change in velocity) / (change in time).
Full Answer
Acceleration is the rate of change of velocity, which is represented by the gradient of the velocity-time graph.
Since the graph is a curve at t=2, we must draw a tangent to the curve at that point to estimate the gradient.
**Step 1: Draw a tangent to the curve at t = 2.**
The point on the curve at t=2 is approximately (2, 5.5). A tangent drawn at this point might pass through, for example, (0, 2.5) and (4, 8.5). (Note: answers will vary depending on the tangent drawn).
**Step 2: Calculate the gradient of the tangent.**
Gradient = (change in y) / (change in x) = (change in velocity) / (change in time)
Using the points (0, 2.5) and (4, 8.5):
Gradient = (8.5 - 2.5) / (4 - 0)
Gradient = 6 / 4
Gradient = 1.5 m/s²
**Answer: Approximately 1.5 m/s²** (Any answer in the range 1.3 to 1.7 would likely be accepted if a suitable tangent is shown).
Acceleration is the rate of change of velocity. On a velocity-time graph, this is represented by the gradient.
At t = 2 seconds, the graph is a curve. To find the gradient at a single point on a curve, we must draw a tangent to the curve at that point. A tangent is a straight line that touches the curve at that point and has the same steepness as the curve at that point.
1. **Locate the point:** Find t=2 on the x-axis. The corresponding point on the curve has a velocity of approximately 5.5 m/s.
2. **Draw a tangent:** Draw a straight line that just touches the curve at (2, 5.5).
3. **Find the gradient of the tangent:** Pick two points on the tangent line that are easy to read and far apart for better accuracy. For example, the tangent might pass through (1, 4) and (4, 8.5).
4. **Calculate the gradient:**
Gradient = (change in y) / (change in x)
= (8.5 - 4) / (4 - 1)
= 4.5 / 3
= 1.5 m/s²
The estimated acceleration after 2 seconds is 1.5 m/s². (Note: Your answer may vary slightly depending on how you draw your tangent, but it should be in a similar range).
Common mistakes
✗ Trying to calculate the gradient by just reading the value at t=2 (e.g., 5.5 / 2), which is incorrect.
✗ Drawing a line that crosses the curve (a secant) instead of a tangent.
✗ Choosing points on the curve itself to calculate the gradient, rather than points on the tangent line.
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