How to Sketch a Position-Time Graph from a Velocity-Time Graph
Learn how to convert a piecewise velocity-time graph into a position-time graph using signed areas, gradients, and curvature. Includes worked examples and challenging HSC-style practice questions with detailed solutions.
A trolley starts at \(x=0\) and its velocity decreases uniformly from \(+4.0\ \mathrm{m\,s^{-1}}\) to \(-2.0\ \mathrm{m\,s^{-1}}\) over 6.0 seconds. What should its position-time graph look like?
You might predict a straight line because the velocity-time graph is a straight line. That prediction is wrong. The trolley initially moves forwards, slows down, stops momentarily, and then moves backwards. Its position-time graph is a curved line that rises to a maximum before falling again.
To sketch a position-time (\(x\)-\(t\)) graph from a piecewise velocity-time (\(v\)-\(t\)) graph, you need two pieces of information:
- The signed area under the velocity-time graph tells you how the object’s position changes.
- The velocity at each instant tells you the gradient of the position-time graph.
For the trolley, velocity reaches zero at \(t=4.0\ \mathrm{s}\). The positive area before this instant is \(8.0\ \mathrm{m}\), while the negative area afterwards is \(2.0\ \mathrm{m}\).
Therefore, the trolley reaches a maximum position of \(8.0\ \mathrm{m}\) at 4.0 seconds and finishes at \(6.0\ \mathrm{m}\) after 6.0 seconds.
Let’s build a reliable method for doing this with different combinations of constant and changing velocity.
01Understand the two jobs: area and gradient
Before sketching anything, remember that the two graphs answer different questions.
A velocity-time graph tells you how quickly an object’s position is changing. A position-time graph tells you where the object is at each instant.
The connection is:
\[ v=\frac{\Delta x}{\Delta t} \]
Here:
- \(v\) is velocity in metres per second (\(\mathrm{m\,s^{-1}}\)).
- \(\Delta x\) is displacement in metres (\(\mathrm{m}\)).
- \(\Delta t\) is the time interval in seconds (\(\mathrm{s}\)).
For instantaneous velocity, the gradient of the tangent to the position-time graph gives the velocity at that instant.
In other words:
\[ \boxed{\text{Gradient of }x\text{-}t\text{ graph}=v} \]
But how do you know where to place the position-time graph vertically?
That’s where area comes in.
Signed area gives displacement
The signed area between a velocity-time graph and the time axis gives displacement.
\[ \Delta x=\text{signed area under the }v\text{-}t\text{ graph} \]
If the velocity is positive, the area contributes positive displacement.
If the velocity is negative, the area contributes negative displacement.
The units confirm the relationship:
\[ \mathrm{\frac{m}{s}}\times\mathrm{s}=\mathrm{m} \]
This is why you can use rectangles, triangles, and trapeziums to calculate displacement from a velocity-time graph.
One important detail: area gives displacement, not automatically the object’s final position.
If the object starts at position \(x_0\), its position at a later time is:
\[ \boxed{x=x_0+\Delta x} \]
For example, if an object starts at \(x=5.0\ \mathrm{m}\) and accumulates a displacement of \(-3.0\ \mathrm{m}\), its final position is \(2.0\ \mathrm{m}\), not \(-3.0\ \mathrm{m}\).
What the different velocity regions tell you
| Feature of the \(v\)-\(t\) graph | What happens on the \(x\)-\(t\) graph? |
|---|---|
| Positive velocity | Position increases; positive gradient |
| Negative velocity | Position decreases; negative gradient |
| Zero velocity | Horizontal tangent |
| Constant positive velocity | Straight line sloping upwards |
| Constant negative velocity | Straight line sloping downwards |
| Increasing velocity | Concave-up curve |
| Decreasing velocity | Concave-down curve |
| Velocity crosses zero | Position reaches a local maximum or minimum |
The last two rows are particularly important.
An increasing velocity means positive acceleration, so the position-time graph curves upwards. This remains true even if the velocity is negative.
Likewise, decreasing velocity means negative acceleration, so the position-time graph curves downwards.
Do not confuse the direction of motion with the direction of acceleration.
02A reliable method for sketching the position-time graph
Use the following process whenever you are given a piecewise velocity-time graph.
Step 1: Identify the initial position
Find the object’s position at \(t=0\).
If the question states that the object starts at \(x=3.0\ \mathrm{m}\), your position-time graph must start at \((0,3.0)\).
If no initial position is supplied, you cannot determine the absolute vertical position of the graph. You can still determine its shape by choosing \(x_0=0\), provided you state that assumption.
Step 2: Mark important times
Identify:
- The start and end of every velocity section.
- Every point where velocity reaches zero.
- Every point where the velocity changes from positive to negative, or vice versa.
These are the times where something important may happen to the position-time graph.
A zero-velocity crossing is particularly important because it can indicate a turning point.
Step 3: Calculate displacement using signed areas
Work from left to right along the velocity-time graph.
Calculate the area of each section using the appropriate geometric formula.
| Shape | Area |
|---|---|
| Rectangle | \(A=bh\) |
| Triangle | \(A=\frac12 bh\) |
| Trapezium | \(A=\frac12(a+b)h\) |
Remember that \(b\) and \(h\) represent the relevant dimensions of the region, not necessarily the same physical quantities in every diagram.
For a velocity-time graph, the horizontal dimension is time and the vertical dimension is velocity.
Areas below the time axis contribute negative displacement.
Step 4: Calculate the position at each important time
Start with the initial position.
Add each positive displacement contribution and subtract each negative contribution.
Build a table as you go.
For example:
| Time (s) | Displacement during interval (m) | Position (m) |
|---|---|---|
| 0 | – | 2 |
| 2 | +6 | 8 |
| 4 | -2 | 6 |
| 6 | -4 | 2 |
This gives you the coordinates that your position-time graph must pass through.
Step 5: Use velocity to determine the gradient
Now consider the shape between your calculated points.
Ask:
- Is velocity positive or negative?
- Is velocity constant or changing?
- Is velocity increasing or decreasing?
- Does velocity reach zero?
For example, if velocity decreases from \(+4\) to \(+2\ \mathrm{m\,s^{-1}}\), the position graph must continue rising, but its gradient becomes less positive.
It should curve downwards rather than remain a straight line.
Step 6: Join the points with the correct shape
Use the velocity information to connect your calculated positions.
A constant velocity produces a straight section.
A linearly changing velocity produces a parabolic section because acceleration is constant over that interval.
If velocity is continuous at a boundary, the position-time graph must have the same gradient immediately before and after that boundary.
If velocity jumps suddenly, the position remains continuous, but the gradient changes abruptly.
This gives you a final check: your graph must satisfy both the area calculations and the gradient information.
03What happens when velocity changes at a boundary?
Piecewise graphs often contain several sections joined together. Understanding what happens at these joins prevents some very common sketching mistakes.
Case 1: Velocity changes, but remains continuous
Imagine velocity decreases from \(+4\) to \(+2\ \mathrm{m\,s^{-1}}\), then continues decreasing at a different rate.
The velocity graph has a sharp corner because its gradient changes.
However, the velocity itself does not jump.
The position-time graph must therefore have the same gradient on both sides of the boundary. Its curvature can change, but it should not have a sharp corner.
This is because velocity determines the gradient, while acceleration determines how that gradient changes.
Case 2: Velocity jumps suddenly
Now imagine velocity changes instantly from \(+3\) to \(-1\ \mathrm{m\,s^{-1}}\).
The position-time graph cannot jump vertically.
The object does not teleport to another position just because its velocity changes.
Instead, its position remains continuous while its gradient changes from positive to negative.
This produces a corner in the idealised position-time graph.
In a real system, a finite force acting over a finite time produces a continuous velocity change. An instantaneous jump is a mathematical idealisation.
Case 3: Velocity reaches zero
This is where you need to pay attention to the sign of velocity on either side of zero.
There are two different possibilities.
Velocity crosses zero: It changes sign, so the object reverses direction. The position-time graph has a local maximum or minimum.
Velocity touches zero: It reaches zero but remains the same sign on either side. The object momentarily stops but does not reverse direction. The position graph has a horizontal tangent without a maximum or minimum.
04Worked examples
The best way to become comfortable with this method is to practise moving between the two graphs.
Worked example: Constant velocity in two directions
An object starts at position \(x=-4.0\ \mathrm{m}\). Its velocity is \(+2.0\ \mathrm{m\,s^{-1}}\) from \(t=0\) to \(t=3.0\ \mathrm{s}\), then changes instantly to \(+1.0\ \mathrm{m\,s^{-1}}\) until \(t=5.0\ \mathrm{s}\).
Calculate its position at \(t=3.0\ \mathrm{s}\) and \(t=5.0\ \mathrm{s}\), then describe the position-time graph.
Step 1
The object travels with constant positive velocity for 3.0 seconds.
The area is a rectangle:
\[ \Delta x_1=2.0\times3.0 =6.0\ \mathrm{m} \]
The position at 3.0 seconds is:
\[ x_1=-4.0+6.0 \]
\[ x_1=+2.0\ \mathrm{m} \]
Step 2
The velocity is now \(+1.0\ \mathrm{m\,s^{-1}}\) for 2.0 seconds.
\[ \Delta x_2=1.0\times2.0 =2.0\ \mathrm{m} \]
Therefore:
\[ x_2=2.0+2.0 \]
\[ x_2=+4.0\ \mathrm{m} \]
Step 3
The velocity is positive throughout, so the position increases throughout.
However, the gradient changes from \(+2.0\) to \(+1.0\ \mathrm{m\,s^{-1}}\) at 3.0 seconds.
The graph therefore has two straight sections, with a sharp change in gradient at the boundary.
Notice that the position does not jump when the velocity changes.
Worked example: A linearly changing velocity that reverses direction
A cart starts at position \(x=-2.0\ \mathrm{m}\).
Its velocity changes linearly from \(+4.0\ \mathrm{m\,s^{-1}}\) at \(t=0\) to \(+2.0\ \mathrm{m\,s^{-1}}\) at \(t=4.0\ \mathrm{s}\).
It then decreases linearly to \(-2.0\ \mathrm{m\,s^{-1}}\) at \(t=8.0\ \mathrm{s}\).
Calculate:
- The position at \(t=4.0\ \mathrm{s}\).
- The time when the cart reaches its maximum position.
- The maximum position.
- The position at \(t=8.0\ \mathrm{s}\).
Step 1
The velocity decreases linearly from \(+4.0\) to \(+2.0\ \mathrm{m\,s^{-1}}\).
The area is a trapezium:
\[ \begin{aligned} \Delta x_1 &=\frac12(4.0+2.0)(4.0)\\ &=12.0\ \mathrm{m} \end{aligned} \]
The position at 4.0 seconds is:
\[ x(4)= -2.0+12.0 \]
\[ x(4)=10.0\ \mathrm{m} \]
Step 2
Between 4.0 and 8.0 seconds, velocity decreases from \(+2.0\) to \(-2.0\ \mathrm{m\,s^{-1}}\).
Because the graph is a straight line, it crosses zero halfway through the interval.
\[ t=4.0+\frac{4.0}{2} \]
\[ t=6.0\ \mathrm{s} \]
The cart reaches zero velocity at 6.0 seconds.
Since velocity changes from positive to negative, this is a maximum position.
Step 3
Between 4.0 and 6.0 seconds, the positive area is a triangle.
\[ \Delta x_2=\frac12(2.0)(2.0) \]
\[ \Delta x_2=2.0\ \mathrm{m} \]
Therefore:
\[ x(6)=10.0+2.0 \]
\[ x_{\max}=12.0\ \mathrm{m} \]
Step 4
Between 6.0 and 8.0 seconds, velocity is negative.
The triangular area contributes negative displacement:
\[ \Delta x_3=-\frac12(2.0)(2.0) \]
\[ \Delta x_3=-2.0\ \mathrm{m} \]
Therefore:
\[ x(8)=12.0-2.0 \]
\[ x(8)=10.0\ \mathrm{m} \]
The cart finishes at the same position it occupied at 4.0 seconds.
This does not mean it remained stationary between those times. It moved forwards, stopped momentarily, and then moved backwards.
Step 5
The velocity decreases throughout the motion, so acceleration is negative throughout.
The position-time graph is therefore concave down.
At 4.0 seconds, velocity is continuous, so the position graph has no corner. Its gradient remains \(+2.0\ \mathrm{m\,s^{-1}}\), although its curvature changes because acceleration becomes more negative.
At 6.0 seconds, the gradient becomes zero. The graph reaches its maximum and then slopes downwards.
05Common mistakes to avoid
Before trying the questions, check that you understand these distinctions.
| Common mistake | Correct approach |
|---|---|
| Copying the shape of the velocity graph directly | Velocity determines the gradient, not the position. |
| Adding every area as positive | Use signed areas to calculate displacement. |
| Forgetting the initial position | Add displacement to the initial position. |
| Drawing a corner whenever velocity changes | A corner occurs only when velocity itself jumps. |
| Treating zero velocity as automatically a maximum | Check whether velocity changes sign. |
| Assuming positive acceleration means increasing position | Acceleration determines curvature; velocity determines whether position increases or decreases. |
| Drawing a straight line for a linearly changing velocity | Constant acceleration produces a curved position-time graph. |
One more distinction is worth remembering.
If a question asks for total distance rather than position change, you must add the magnitudes of the areas instead of their signed values. The method is explained in How to Calculate Distance When a Velocity-Time Graph Crosses Zero.
06Questions and solutions
The following questions build from straightforward area calculations to more challenging situations involving changing acceleration, stationary points, and repeated returns to the starting position.
Assume one-dimensional motion throughout, with positive velocity representing motion in the positive direction.
Question 1
A trolley begins at position \(x=1.0\ \mathrm{m}\).
Its velocity is:
- \(+3.0\ \mathrm{m\,s^{-1}}\) from \(t=0\) to \(t=2.0\ \mathrm{s}\).
- \(-1.0\ \mathrm{m\,s^{-1}}\) from \(t=2.0\) to \(t=5.0\ \mathrm{s}\).
Assume the velocity changes instantaneously at 2.0 seconds.
Calculate:
a) The position at \(t=2.0\ \mathrm{s}\).
b) The position at \(t=5.0\ \mathrm{s}\).
c) Describe the shape of the position-time graph, including what happens at 2.0 seconds.
Solution 1
The trolley is at \(x=7.0\ \mathrm{m}\) after 2.0 seconds and \(x=4.0\ \mathrm{m}\) after 5.0 seconds. Its position-time graph rises initially, then falls, with a sharp corner at 2.0 seconds.
Step 1: Calculate the displacement during the first interval.
\[ \Delta x_1=3.0\times2.0 =6.0\ \mathrm{m} \]
Add this to the initial position:
\[ x(2)=1.0+6.0 \]
\[ x(2)=7.0\ \mathrm{m} \]
Step 2: Calculate the displacement during the second interval.
The velocity is negative, so the displacement is negative.
\[ \Delta x_2=(-1.0)(5.0-2.0) \]
\[ \Delta x_2=-3.0\ \mathrm{m} \]
Therefore:
\[ x(5)=7.0-3.0 \]
\[ x(5)=4.0\ \mathrm{m} \]
Step 3: Describe the graph.
From 0 to 2.0 seconds, the positive velocity produces a rising straight line with gradient \(+3.0\ \mathrm{m\,s^{-1}}\).
After 2.0 seconds, the negative velocity produces a falling straight line with gradient \(-1.0\ \mathrm{m\,s^{-1}}\).
Because velocity changes instantaneously, the gradient changes abruptly. However, position remains continuous at \(x=7.0\ \mathrm{m}\).
Question 2
A cart starts at position \(x=-2.0\ \mathrm{m}\).
Its velocity-time graph consists of two straight-line sections joining these points:
| Time (s) | Velocity (\(\mathrm{m\,s^{-1}}\)) |
|---|---|
| 0 | +5 |
| 4 | +1 |
| 6 | -3 |
Calculate:
a) The position at \(t=4.0\ \mathrm{s}\).
b) The time when the cart reaches its maximum position.
c) The maximum position.
d) The final position at \(t=6.0\ \mathrm{s}\).
Explain why the position-time graph has no corner at \(t=4.0\ \mathrm{s}\).
Solution 2
The cart reaches a maximum position of \(10.25\ \mathrm{m}\) at \(t=4.5\ \mathrm{s}\), and its final position is \(8.0\ \mathrm{m}\).
Step 1: Calculate the displacement from 0 to 4.0 seconds.
The first section is a trapezium.
\[ \begin{aligned} \Delta x_1 &=\frac12(5+1)(4)\\ &=12\ \mathrm{m} \end{aligned} \]
Therefore:
\[ x(4)=-2+12 \]
\[ x(4)=10\ \mathrm{m} \]
Step 2: Find when velocity reaches zero.
Between 4.0 and 6.0 seconds, velocity changes from \(+1\) to \(-3\ \mathrm{m\,s^{-1}}\).
The velocity decreases by \(4\ \mathrm{m\,s^{-1}}\) over 2.0 seconds.
Its rate of change is:
\[ a=\frac{-3-1}{6-4} =-2.0\ \mathrm{m\,s^{-2}} \]
It takes 0.5 seconds to decrease from \(+1\) to zero.
Therefore:
\[ t=4.0+0.5 \]
\[ t=4.5\ \mathrm{s} \]
Velocity changes from positive to negative, so this is the time of maximum position.
Step 3: Calculate the maximum position.
The positive area between 4.0 and 4.5 seconds is a triangle.
\[ \Delta x_2=\frac12(0.5)(1) \]
\[ \Delta x_2=0.25\ \mathrm{m} \]
Therefore:
\[ x_{\max}=10+0.25 \]
\[ x_{\max}=10.25\ \mathrm{m} \]
Step 4: Calculate the final position.
From 4.5 to 6.0 seconds, velocity is negative.
The triangular area is:
\[ \Delta x_3=-\frac12(1.5)(3) \]
\[ \Delta x_3=-2.25\ \mathrm{m} \]
Therefore:
\[ x(6)=10.25-2.25 \]
\[ x(6)=8.0\ \mathrm{m} \]
Step 5: Explain the graph at 4.0 seconds.
There is no corner because velocity is continuous at this point.
The gradient of the position-time graph is \(+1.0\ \mathrm{m\,s^{-1}}\) immediately before and after 4.0 seconds.
However, the acceleration changes from \(-1.0\) to \(-2.0\ \mathrm{m\,s^{-2}}\).
The position-time graph therefore changes curvature while keeping the same tangent gradient.
Question 3
An object starts at position \(x=-1.0\ \mathrm{m}\).
Its velocity-time graph consists of two straight-line sections joining:
| Time (s) | Velocity (\(\mathrm{m\,s^{-1}}\)) |
|---|---|
| 0 | +4 |
| 2 | 0 |
| 4 | +4 |
Calculate the position at \(t=2.0\ \mathrm{s}\) and \(t=4.0\ \mathrm{s}\).
A student claims that the object reaches its maximum position at \(t=2.0\ \mathrm{s}\), because its velocity is zero at that instant.
Is the student correct? Explain the shape of the position-time graph, including its gradient and curvature around \(t=2.0\ \mathrm{s}\).
Solution 3
The positions are \(x(2)=3.0\ \mathrm{m}\) and \(x(4)=7.0\ \mathrm{m}\). The student is incorrect: the object does not reach a maximum position at 2.0 seconds. Its position continues increasing, although its instantaneous gradient becomes zero at that instant.
Step 1: Calculate the displacement from 0 to 2.0 seconds.
The area is a positive triangle:
\[ \Delta x_1=\frac12(2)(4) \]
\[ \Delta x_1=4.0\ \mathrm{m} \]
Therefore:
\[ x(2)=-1.0+4.0 \]
\[ x(2)=3.0\ \mathrm{m} \]
Step 2: Calculate the displacement from 2.0 to 4.0 seconds.
The velocity is positive again.
The second triangular area is also positive:
\[ \Delta x_2=\frac12(2)(4) \]
\[ \Delta x_2=4.0\ \mathrm{m} \]
Therefore:
\[ x(4)=3.0+4.0 \]
\[ x(4)=7.0\ \mathrm{m} \]
Step 3: Determine whether the object reaches a maximum.
The velocity reaches zero at 2.0 seconds, but it remains positive on both sides.
This means the object momentarily stops but does not reverse direction.
The position-time graph continues increasing before and after 2.0 seconds.
Its gradient is zero at exactly \(t=2.0\ \mathrm{s}\), but the gradient is positive immediately before and afterwards.
Step 4: Determine the curvature.
Before 2.0 seconds, velocity decreases, so acceleration is negative.
The position-time graph is concave down.
After 2.0 seconds, velocity increases, so acceleration is positive.
The position-time graph becomes concave up.
The result is a stationary point of inflection: a horizontal tangent where the curve changes concavity, without reaching a maximum or minimum.
This is an important distinction. A zero gradient alone does not prove that a point is a maximum or minimum.
Question 4
An object starts at position \(x=3.0\ \mathrm{m}\).
Its velocity-time graph consists of straight-line sections joining the following points:
| Time (s) | Velocity (\(\mathrm{m\,s^{-1}}\)) |
|---|---|
| 0 | -4 |
| 2 | -2 |
| 5 | +4 |
| 6 | +4 |
Calculate:
a) The position at \(t=2.0\ \mathrm{s}\).
b) The time when the object reaches its minimum position.
c) The minimum position.
d) The position at \(t=5.0\ \mathrm{s}\) and \(t=6.0\ \mathrm{s}\).
Explain why the position-time graph initially decreases even though acceleration is positive.
Solution 4
The object reaches its minimum position of \(-4.0\ \mathrm{m}\) at \(t=3.0\ \mathrm{s}\). Its positions at 2.0, 5.0, and 6.0 seconds are \(-3.0\ \mathrm{m}\), \(0\ \mathrm{m}\), and \(+4.0\ \mathrm{m}\), respectively.
Step 1: Calculate the position at 2.0 seconds.
The first section is a trapezium below the time axis.
\[ \begin{aligned} \Delta x_1 &=\frac12(-4-2)(2)\\ &=-6.0\ \mathrm{m} \end{aligned} \]
Therefore:
\[ x(2)=3.0-6.0 \]
\[ x(2)=-3.0\ \mathrm{m} \]
Step 2: Find when velocity reaches zero.
Between 2.0 and 5.0 seconds, velocity increases linearly from \(-2\) to \(+4\ \mathrm{m\,s^{-1}}\).
The acceleration is:
\[ a=\frac{4-(-2)}{5-2} \]
\[ a=2.0\ \mathrm{m\,s^{-2}} \]
It takes 1.0 second for velocity to increase from \(-2\) to zero.
Therefore:
\[ t=2.0+1.0 \]
\[ t=3.0\ \mathrm{s} \]
Velocity changes from negative to positive, so the object reaches a minimum position.
Step 3: Calculate the minimum position.
From 2.0 to 3.0 seconds, the area is a negative triangle:
\[ \Delta x_2=-\frac12(1)(2) \]
\[ \Delta x_2=-1.0\ \mathrm{m} \]
Therefore:
\[ x_{\min}=-3.0-1.0 \]
\[ x_{\min}=-4.0\ \mathrm{m} \]
Step 4: Calculate the position at 5.0 seconds.
From 3.0 to 5.0 seconds, the velocity is positive.
The triangular area is:
\[ \Delta x_3=\frac12(2)(4) \]
\[ \Delta x_3=4.0\ \mathrm{m} \]
Therefore:
\[ x(5)=-4.0+4.0 \]
\[ x(5)=0\ \mathrm{m} \]
Step 5: Calculate the position at 6.0 seconds.
The velocity remains constant at \(+4.0\ \mathrm{m\,s^{-1}}\).
The displacement is:
\[ \Delta x_4=4.0\times1.0 \]
\[ \Delta x_4=4.0\ \mathrm{m} \]
Therefore:
\[ x(6)=0+4.0 \]
\[ x(6)=4.0\ \mathrm{m} \]
Step 6: Explain the curvature.
This is the subtle part.
From 0 to 5.0 seconds, velocity increases continuously. Therefore, acceleration is positive and the position-time graph is concave up.
However, the object initially has negative velocity.
It is moving in the negative direction, so its position decreases even while acceleration is positive.
At 3.0 seconds, velocity reaches zero and changes sign. The position reaches a minimum, then starts increasing.
After 5.0 seconds, acceleration becomes zero because velocity is constant. The position-time graph becomes a straight line with a positive gradient.
The key distinction is that acceleration determines the curvature, while velocity determines the direction of the position change.
Question 5
An object begins at the origin, so \(x(0)=0\).
Its velocity-time graph has three sections:
- From \(t=0\) to \(t=6.0\ \mathrm{s}\), velocity decreases linearly from \(+4.0\) to \(-2.0\ \mathrm{m\,s^{-1}}\).
- From \(t=6.0\) to \(t=9.0\ \mathrm{s}\), velocity remains constant at \(-2.0\ \mathrm{m\,s^{-1}}\).
- From \(t=9.0\) to \(t=13.0\ \mathrm{s}\), velocity increases linearly from \(-2.0\) to \(+2.0\ \mathrm{m\,s^{-1}}\).
Assume velocity is continuous at both boundaries.
Answer the following:
a) Determine all times after \(t=0\) when the object returns to the origin.
b) Find the times and positions of all local maximum and minimum points on the position-time graph.
c) Explain why returning to the origin does not necessarily mean the object has stopped.
d) Sketch the complete position-time graph, including its curvature and gradients at the important points.
Solution 5
The object returns to the origin at \(t=9.0\ \mathrm{s}\) and \(t=13.0\ \mathrm{s}\). Its local maximum is \(x=8.0\ \mathrm{m}\) at \(t=4.0\ \mathrm{s}\), and its local minimum is \(x=-2.0\ \mathrm{m}\) at \(t=11.0\ \mathrm{s}\).
The important distinction is that returning to the origin and stopping are two different events.
Step 1: Find the first time velocity reaches zero.
During the first interval, velocity decreases linearly from \(+4.0\) to \(-2.0\ \mathrm{m\,s^{-1}}\) over 6.0 seconds.
Its acceleration is:
\[ a=\frac{-2-4}{6} \]
\[ a=-1.0\ \mathrm{m\,s^{-2}} \]
The velocity reaches zero after 4.0 seconds.
Therefore, the object reaches a local maximum position at:
\[ t=4.0\ \mathrm{s} \]
Step 2: Calculate the maximum position.
From 0 to 4.0 seconds, the area is a positive triangle.
\[ \Delta x_1=\frac12(4)(4) \]
\[ \Delta x_1=8.0\ \mathrm{m} \]
Since the object starts at the origin:
\[ x(4)=8.0\ \mathrm{m} \]
This is the maximum position.
Step 3: Calculate the position at 6.0 seconds.
From 4.0 to 6.0 seconds, velocity is negative.
The area is a negative triangle:
\[ \Delta x_2=-\frac12(2)(2) \]
\[ \Delta x_2=-2.0\ \mathrm{m} \]
Therefore:
\[ x(6)=8.0-2.0 \]
\[ x(6)=6.0\ \mathrm{m} \]
Step 4: Calculate the position at 9.0 seconds.
Velocity is constant at \(-2.0\ \mathrm{m\,s^{-1}}\) from 6.0 to 9.0 seconds.
The displacement is:
\[ \Delta x_3=(-2.0)(3.0) \]
\[ \Delta x_3=-6.0\ \mathrm{m} \]
Therefore:
\[ x(9)=6.0-6.0 \]
\[ x(9)=0\ \mathrm{m} \]
The object has returned to the origin.
However, its velocity is still \(-2.0\ \mathrm{m\,s^{-1}}\).
It is moving in the negative direction, so it passes through the origin rather than stopping there.
This is the first important surprise in the question.
Step 5: Find the next time velocity reaches zero.
From 9.0 to 13.0 seconds, velocity increases linearly from \(-2.0\) to \(+2.0\ \mathrm{m\,s^{-1}}\).
The change is symmetric, so velocity reaches zero halfway through the interval.
\[ t=9.0+\frac{4.0}{2} \]
\[ t=11.0\ \mathrm{s} \]
Velocity changes from negative to positive.
Therefore, the object reaches a local minimum position at 11.0 seconds.
Step 6: Calculate the minimum position.
From 9.0 to 11.0 seconds, the area is a negative triangle:
\[ \Delta x_4=-\frac12(2)(2) \]
\[ \Delta x_4=-2.0\ \mathrm{m} \]
Therefore:
\[ x(11)=0-2.0 \]
\[ x_{\min}=-2.0\ \mathrm{m} \]
Step 7: Calculate the position at 13.0 seconds.
From 11.0 to 13.0 seconds, velocity is positive.
The area is a positive triangle:
\[ \Delta x_5=\frac12(2)(2) \]
\[ \Delta x_5=2.0\ \mathrm{m} \]
Therefore:
\[ x(13)=-2.0+2.0 \]
\[ x(13)=0\ \mathrm{m} \]
The object returns to the origin for a second time.
Its velocity at this instant is \(+2.0\ \mathrm{m\,s^{-1}}\), so it is moving in the positive direction.
Step 8: Put the complete position history together.
The important coordinates are:
| Time (s) | Position (m) | Interpretation |
|---|---|---|
| 0 | 0 | Initial position |
| 4 | +8 | Local maximum |
| 6 | +6 | Velocity becomes constant |
| 9 | 0 | First return to origin |
| 11 | -2 | Local minimum |
| 13 | 0 | Second return to origin |
The position-time graph has three different curvature sections.
From 0 to 6 seconds, velocity decreases, so the graph is concave down.
From 6 to 9 seconds, velocity is constant, so the graph is a straight line sloping downwards.
From 9 to 13 seconds, velocity increases, so the graph is concave up.
At 6.0 and 9.0 seconds, velocity remains continuous, so the position-time graph has no corners at either boundary.
The tempting approach is to calculate the total signed area from 0 to 13 seconds, obtain zero, and conclude that the object returns to the origin at 13 seconds.
That calculation is correct, but it misses the earlier return at 9 seconds.
The position at any instant depends on the accumulated signed area up to that instant, not just the total area at the end of the motion.
There is also a second distinction: a zero position does not mean zero velocity. The object can pass through the origin while moving, just as it does at both 9.0 and 13.0 seconds.
This is why you must track position and velocity separately when constructing the graph.
07What to remember when sketching these graphs
When converting a piecewise velocity-time graph into a position-time graph, keep two questions in mind:
- Where is the object? Calculate this using the initial position and the accumulated signed area.
- How is its position changing? Use velocity to determine the gradient and acceleration to determine the curvature.
If velocity is constant, draw a straight line.
If velocity changes linearly, draw a parabolic section.
If velocity crosses zero, check for a maximum or minimum. If it only touches zero, check whether the object continues in the same direction.
And always check that your final position agrees with the total signed area.
Once you can sketch position-time graphs from velocity-time graphs, the next useful skill is working backwards: using the gradient of a position-time graph to determine velocity, and then using changes in that gradient to determine acceleration. These connections bring the three main motion graphs together. For the relationship between acceleration and velocity, see Acceleration-Time Graphs: Area and Velocity Change.