Graphs of motion are crucial for physicists to determine the position and speed of an object. Unlike maps and speedometers which are useful for non-physicists, graphs provide a detailed analysis of an object's movement over time. As a physics student, you'll come to realize the importance of graphs of motion in understanding the movement of a body. In short, these graphs help us determine the rate of change of an object's speed and its position at any given time.

There are three main types of graphs used to define the motion of an object in a straight line: displacement-time graphs, velocity-time graphs, and acceleration-time graphs.

Figure 1 illustrates a displacement-time graph of an object moving at a constant velocity. For the displacement-time graph, displacement (denoted by d) is on the y-axis, and time (denoted by t) is on the x-axis.

Graphs of motion provide us with valuable information about an object's movement. By looking at the graph, we can calculate the distance covered at any given time, the average velocity by finding the slope of the graph, and the instantaneous velocity by calculating the derivative of any point on the curve. To calculate the slope (p) of a displacement-time graph, we use the equation: [insert equation here]. Basically, the slope of the displacement-time graph gives us the velocity because velocity is the rate of change of displacement.

Have a look at the velocity-time graph below:

The velocity-time graph is another useful tool for physicists to understand the movement of an object. Velocity (v) is on the y-axis and time (t) is on the x-axis. From this graph, we can determine the velocity of the object at any given time, the average acceleration by finding the slope of the straight line, the instantaneous acceleration by calculating the derivative of any point on the curve, and the displacement of the object by calculating the area under the curve (between the line and the time axis). To calculate the slope (p) of the velocity-time graph, we use the equation: [insert equation here]. The slope of the velocity-time graph gives us the acceleration because acceleration is the rate of change of velocity. Additionally, the area under the velocity-time graph gives us the distance covered by the object, which is the displacement.

For an acceleration-time graph, acceleration (a) is given on the y-axis and the time (t) on the x-axis. The acceleration-time graph gives us the acceleration at any given time. Also, the area under the acceleration-time curve represents a change in velocity.

Below we explore how to draw graphs of motion for different scenarios.

For an object at rest, there will be no change in displacement, which will result in no change in velocity, and because there is no change in the velocity, the change in acceleration will be zero as well. An object at rest will not move. Hence, the displacement will not change over the interval of time which is depicted by a flat line parallel to the time axis.

The velocity will be zero because the object’s displacement doesn’t change. Hence, the graph for an object without its velocity changing over time can be shown with a straight line on the time axis.

The acceleration will be zero because the object’s velocity is not changing and the acceleration time graph be a flat line starting from the origin.

When an object moves at a constant velocity, the rate of change of displacement will be uniformly increasing. The velocity will be a continuous straight line with a positive or negative value depending on the direction of movement. Acceleration will be zero, so a constant line will pass through the origin. In terms of the graph, if the slope is positive, it indicates that the movement is in the positive direction (away from the origin). If the slope had been negative (towards the origin), it would have depicted displacement in the opposite direction. Additionally, the displacement will be uniformly increasing because the velocity is constant.

Question: Which direction should be considered positive or negative? Answer: The sign is arbitrary. You can take any direction as either positive or negative.

As the slope of the displacement-time graph for a body moving with a constant velocity is positive in figure 7, the velocity is a constant straight line in the positive direction.

As there is no rate of change of velocity (constant velocity), the acceleration will be zero as well because for acceleration or deceleration to occur, there needs to be a change in velocity as well.

When an object moves with constant acceleration:

The displacement-time graph will be a curve with a gradient that becomes increasingly steeper. The rate of change of velocity will be uniformly increasing. Acceleration will be constant with a positive value.

Below are the two graphs for displacement vs time. Figure 10 is for constant acceleration and figure 11 is for constant deceleration.

If you take the tangents at various points on both of the above curves, you will see that the slope of the displacement-time graph in figure 10 is becoming steeper and steeper. This is an indication that the velocity is increasing. In figure 11, the gradient is gradually decreasing, which is an indication that the velocity is decreasing.

The velocity-time graph for a constant acceleration will be a uniformly increasing line as shown by the figure below.

As the acceleration is not changing over time and is constant, the acceleration-time graph can be represented by a straight line.

When an object moves with a constant deceleration:

The displacement-time graph will be a curve with a gradient that becomes more and more horizontal. The rate of change of velocity will uniformly decrease. Acceleration will be constant but with a negative value. Because the body is decelerating, the curve is approaching a constant (unchanged) value.

The velocity-time graph for a constant deceleration will be a uniform line constantly decreasing from some value.

The constant line with a negative acceleration shows that an object is decelerating with a constant value.

In this scenario, the ball is thrown upwards with an initial velocity (v1) of 10 meters per second. The acceleration due to gravity (g) is 9.81 meters per second squared downward. This means that the ball will accelerate downwards until it reaches its maximum height, and then it will begin to decelerate until it reaches the thrower's hand.

The ball's position (y) and velocity (v) can be calculated using the equations of motion, which state that the position of the ball at any given time (t) is equal to the initial position (y1) plus the initial velocity (v1) multiplied by the time (t) plus one half of the acceleration (g) multiplied by the square of the time (t2). This equation can be written as: y = y1 + v1t + (1/2)gt2.

The velocity of the ball at any given time (t) is equal to the initial velocity (v1) plus the acceleration (g) multiplied by the time (t). This equation can be written as: v = v1 + gt.

Using these equations, we can calculate the position and velocity of the ball at any given time. For example, at 1.00 s the ball is above its starting point and heading upward, since y1 and v1 are both positive. At 2.00 s, the ball is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both y3 and v3 are negative, meaning the ball is below its starting point and continuing to move downward.

The ball will eventually reach its maximum height, which is equal to the initial position (y1) plus the initial velocity (v1) multiplied by the time (t) plus one half of the acceleration (g) multiplied by the square of the time (t2). This equation can be written as: ymax = y1 + v1t + (1/2)gt2.

Once the ball reaches its maximum height, it will begin to decelerate and eventually come to rest in the thrower's hand. The time it takes for the ball to reach its maximum height and come to rest in the thrower's hand can be calculated using the equation of motion. This equation can be written as: t = (2v1)/g.

Using this equation, we can calculate that it will take 0.45173 s for the ball to reach its maximum height and come to rest in the thrower's hand. The ball will be 1.0000 m above its initial position when it comes to rest.

The displacement-time graph for an object thrown straight up and then landing in the thrower's hand is shown below.

You are correct that the ball's displacement increases as it is thrown upwards, and that the displacement-time graph will be a curve rather than a straight line.

The reason for the curve is due to the acceleration due to gravity. As the ball is thrown upwards, it is moving against the force of gravity, which is pulling it downwards. As a result, the ball's velocity decreases until it reaches its maximum height, at which point its velocity becomes zero.

After reaching its maximum height, the ball begins to fall back down towards the ground, with the acceleration due to gravity acting upon it. This acceleration due to gravity is a constant force, and it causes the ball's velocity to increase as it falls back towards the ground.

Because the velocity is changing at a constant rate due to the acceleration, the displacement-time graph will be curved. The curve is a result of the fact that the acceleration due to gravity is a constant force acting on the ball, causing its velocity to change at a constant rate.

Additionally, the motion of the ball is not linear, but rather a two-dimensional motion. This is because the ball is moving in both the vertical and horizontal directions. The curve on the displacement-time graph represents the motion of the ball in the vertical direction only, while the horizontal motion remains constant.

In conclusion, the curve on the displacement-time graph is a result of the constant acceleration due to gravity acting on the ball as it is thrown upwards and then falls back down towards the ground.

The velocity-time graph for an object thrown straight up and then landing in the thrower's hand is shown below.

The ball is thrown upwards with some initial velocity u. As the ball reaches the top, its velocity decreases uniformly until it reaches zero, where the ball is at rest for a brief moment. Afterwards, the ball moves downwards with a uniformly increasing velocity. As the distance travelled will be the same upwards and downwards because of negligible air resistance, the initial velocity will be equal to the final velocity -u. So, the area of both regions A and B will be the same in this case. Why is the slope of the graph negative and not positive after u reaches zero? As the upwards direction is taken as positive, once the direction of the ball changes at the top, the motion will be downwards in the negative direction with a constant acceleration of free fall.

The acceleration-time graph for an object thrown straight up and then landing in the thrower's hand is shown below.

The acceleration is a constant -9.81m/s2 throughout the displacement as the velocity-time graph is uniformly decreasing. After the ball is tossed in the air, the gravitational force works in the direction opposite to the upwards motion. Since the upwards motion is taken to be positive, the gravitational force will be negative. Once the ball reaches its peak, the ball changes direction. Hence, the gravitational force will continue to be negative.

You are correct that the three main types of graphs concerning linear motion are the displacement-time graph, the velocity-time graph, and the acceleration graph.

From the displacement-time graph, you can calculate the average velocity by finding the slope of the graph, which is equal to the change in displacement over the change in time. You can also calculate the instantaneous by finding the slope of the tangent line to the graph at a specific point in time.

From the velocity-time graph, you can calculate the average acceleration by finding the slope of the graph, which is equal to the change in velocity over the change in time. You can also calculate the instantaneous acceleration by finding the slope of the tangent line to the graph at a specific point in time.

The area under the velocity-time graph represents displacement, which is the total distance travelled by an object. This is because the area under the graph represents the product of velocity and time, which is equal to displacement.

The area under the acceleration-time graph represents a change in velocity, which is the total change in velocity experienced by an object. This is because the area under the graph represents the product of acceleration and time, which is equal to the change in velocity.

In conclusion, the three main types of graphs concerning linear motion can provide important information about an object's motion, including its velocity, acceleration, and displacement. Understanding how to interpret these graphs is important for analyzing the motion of objects in both physics and engineering.

**How to interpret graphs of motion?**

You can interpet graphs of motion as follows: The gradient of the graph and area under the graph give information about the displacement, velocity, and acceleration depending on what type of linear motion graph we are considering. For example, the slope of a displacement-time graph gives the average velocity.

**What are the graphs of motion?**

The graphs of motion are the displacement-time graph, the velocity-time graph, and the acceleration-time graph.

**What type of graph is used to display motion?**

Graphs of motion are used to display linear motion. The graphs of motion are the displacement-time graph, the velocity-time graph, and the acceleration-time graph.

**How to work out velocity of motion graphs?**

You can work out the velocity of motion graphs as follows: You can find the average velocity by calculating the slope of the displacement-time graph. You can find the velocity at any given time on a velocity-time graph. You can find the velocity by calculating the area under the acceleration-time curve from an acceleration-time graph.

for Free

14-day free trial. Cancel anytime.

Join **20,000+** learners worldwide.

The first 14 days are on us

96% of learners report x2 faster learning

Free hands-on onboarding & support

Cancel Anytime