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To find the displacement from the velocity-time graph. you need to find the area of the shape and the shape is under the line

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drag races and dune climbing races

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1a. 625m

1b. 122.5m

2a. 240miles

2b. 15m

2c.0m

3a. 70m

3b. 308m

3c. 378m

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V1: How to calculate displacement from v-t graph

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V1: How to calculate displacement from v-t graph

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Find quizzes on any topic and practice or compete with friends.

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As learned in an earlier part of this lesson, a plot of velocity-time can be used to determine the acceleration of an object (the slope). In this part of the lesson, we will learn how a plot of velocity versus time can also be used to determine the displacement of an object. For velocity versus time graphs, the area bound by the line and the axes represents the displacement. The diagram below shows three different velocity-time graphs; the shaded regions between the line and the time-axis represent the displacement during the stated time interval.

The shaded area is representative of the displacement during from 0 seconds to 6 seconds. This area takes on the shape of a rectangle can be calculated using the appropriate equation.

The shaded area is representative of the displacement during from 0 seconds to 4 seconds. This area takes on the shape of a triangle can be calculated using the appropriate equation.

The shaded area is representative of the displacement during from 2 seconds to 5 seconds. This area takes on the shape of a trapezoid can be calculated using the appropriate equation.

The method used to find the area under a line on a velocity-time graph depends upon whether the section bound by the line and the axes is a rectangle, a triangle or a trapezoid. Area formulas for each shape are given below.

Rectangle

Triangle

Trapezoid

Area = b * h

Area = ½ * b * h

Area = ½ * b * (h ₁ + h ₂)

Calculating the Area of a Rectangle

Now we will look at a few example computations of the area for each of the above geometric shapes. First consider the calculation of the area for a few rectangles. The solution for finding the area is shown for the first example below. The shaded rectangle on the velocity-time graph has a base of 6 s and a height of 30 m/s. Since the area of a rectangle is found by using the formula A = b x h, the area is 180 m (6 s x 30 m/s). That is, the object was displaced 180 meters during the first 6 seconds of motion.

Area = b * h
Area = (6 s) * (30 m/s)

Area = 180 m

Now try the following two practice problems as a check of your understanding. Determine the displacement (i.e., the area) of the object during the first 4 seconds (Practice A) and from 3 to 6 seconds (Practice B).

Calculating the Area of a Triangle

Now we will look at a few example computations of the area for a few triangles. The solution for finding the area is shown for the first example below. The shaded triangle on the velocity-time graph has a base of 4 seconds and a height of 40 m/s. Since the area of triangle is found by using the formula A = ½ * b * h, the area is ½ * (4 s) * (40 m/s) = 80 m. That is, the object was displaced 80 meters during the four seconds of motion.

Area = ½ * b * h
Area = ½ * (4 s) * (40 m/s)

Area = 80 m

Now try the following two practice problems as a check of your understanding. Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B).

Calculating the Area of a Trapezoid

Finally we will look at a few example computations of the area for a few trapezoids. The solution for finding the area is shown for the first example below. The shaded trapezoid on the velocity-time graph has a base of 2 seconds and heights of 10 m/s (on the left side) and 30 m/s (on the right side). Since the area of trapezoid is found by using the formula A = ½ * (b) * (h ₁ + h ₂), the area is 40 m [½ * (2 s) * (10 m/s + 30 m/s)]. That is, the object was displaced 40 meters during the time interval from 1 second to 3 seconds.

Area = ½ * b * (h ₁ + h ₂)
Area = ½ * (2 s) * (10 m/s + 30 m/s)

Area = 40 m

Now try the following two practice problems as a check of your understanding. Determine the displacement of the object during the time interval from 2 to 3 seconds (Practice A) and during the first 2 seconds (Practice B).

Alternative Method for Trapezoids

An alternative means of determining the area of a trapezoid involves breaking the trapezoid into a triangle and a rectangle. The areas of the triangle and rectangle can be computed individually; the area of the trapezoid is then the sum of the areas of the triangle and the rectangle. This method is illustrated in the graphic below.

Triangle: Area = ½ * (2 s) * (20 m/s) = 20 m

Rectangle: Area = (2 s) * (10 m/s) = 20 m

Total Area = 20 m + 20 m = 40 m

It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. The area can be identified as a rectangle, triangle, or trapezoid. The area can be subsequently determined using the appropriate formula. Once calculated, this area represents the displacement of the object.

Investigate!
The widget below computes the area between the line on a velocity-time plot and the axes of the plot. This area is the displacement of the object. Use the widget to explore or simply to practice a few self-made problems.

We Would Like to Suggest ...

Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Two Stage Rocket Interactive. This Interactive is found in the Physics Interactives section of our website and allows a learner to apply the skill of calculating areas and relating them to displacement values for a two-stage rocket.

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what does the slope of a v-t graph represent

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Velocity-Time Graphs: Determining the Area (and Displacement)

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Kinematics is the science of describing the motion of objects. One method for describing the motion of an object is through the use of velocity-time graphs which show the velocity of the object as a function of time. The area between the line and the time-axis on these graphs is equal to the displacement of the object.

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V3: Motion with Constant acceleration equations

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V3: Motion with Constant acceleration equations

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Examples on how to use Kinematic equations

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Examples on how to use Kinematic equations

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Earlier in Lesson 6, four kinematic equations were introduced and discussed. A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented. These problems allow any student of physics to test their understanding of the use of the four kinematic equations to solve problems involving the one-dimensional motion of objects. You are encouraged to read each problem and practice the use of the strategy in the solution of the problem. Then click the button to check the answer or use the link to view the solution.

Check Your Understanding

1. An airplane accelerates down a runway at 3.20 m/s² for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

2. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s². Determine the time for the feather to fall to the surface of the moon.

3. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels?

4. An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s². The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

5. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)

6. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

7. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s², determine the speed of the Jaguar before it began to skid.

8. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

Solutions to Above Problems

1. a = (Delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s ²

d = v i*t + 0.5*a*t ²

d = 45.7 m + 34.1 m

d = 79.8 m

(Note: the d can also be calculated using the equation v f ² = v i ² + 2*a*d)
Return to Problem 4

2. a = (Delta v)/t
a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s ²

d = v i*t + 0.5*a*t ²

d = 0 m + 406 m

d = 406 m

(Note: the d can also be calculated using the equation v f ² = v i ² + 2*a*d)
Return to Problem 6

3. (0 m/s) ² = v i ² + 2*(-9.8 m/s ²)*(1.29 m)

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

v f = v i + a*t

0 m/s = 5.03 m/s + (-9.8 m/s ²)*t up

-5.03 m/s = (-9.8 m/s ²)*t up

(-5.03 m/s)/(-9.8 m/s ²) = t up

t up = 0.513 s

hang time = 1.03 s
Return to Problem 11

(NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use: v f = v i + a*t

0 m/s = v i + (-9.8 m/s ²)*(3.13 s)

0 m/s = v i - 30.7 m/s

v i = 30.7 m/s (30.674 m/s)

Now use: v f ² = v i ² + 2*a*d

(0 m/s) ² = (30.7 m/s) ² + 2*(-9.8 m/s ²)*(d)

0 m ²/s ² = (940 m 2 /s 2 ) + (-19.6 m/s ²)*d

-940 m ²/s ² = (-19.6 m/s ²)*d

(-940 m ²/s ²)/(-19.6 m/s ²) = d

d = 48.0 m
Return to Problem 13

5. d = vi*t + 0.5*a*t² Return to Problem 14
t = 8.69 s

6. d = vi*t + 0.5*a*t²

d = -57.0 m

(NOTE: the - sign indicates direction)
Return to Problem 16

7. First, find speed in units of m/s:

v f ² = v i ² + 2*a*d

(0 m/s) ² = v i ² + 2*(-9.8 m/s ²)*(91.5 m)

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr
Return to Problem 20

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Kinematic equations relate the variables of motion to one another. Each equation contains four variables. The variables include acceleration (a), time (t), displacement (d), final velocity (vf), and initial velocity (vi). If values of three variables are known, then the others can be calculated using the equations.

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This lesson starts at page 65

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On a velocity-time graph, the area under the line represents the object's displacement.

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The area under the line depicts the objects displacement.

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A drag racing.

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A motion in which the speed increases at the same rate every second.

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to calculate the final speed of a drag racing car

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The distance travelled by an object can be calculated from the areaunder a velocity-time graph.

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drag race because its move in straight line.

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Drag race

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the area is under the line of depicts and the displacement of objects

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khalifa

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The area under the line on a velocity-time graph is equal to the displacement of the object.

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xf - xi = vxi∆t + ½ax∆t²

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The motion in which increases every second at a constant rate.

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To find the displacement from the velocity-time graph, you need to find the area of the shape. The shape is gotten from under the line.

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To calculate Δx= Area

Rectangle= Area = b • h

Triangle= Area = ½ • b • h

Trapezoid=Area = ½ • b • (h₁ + h₂)

Find the area under the line.

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A golf ball.

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the area under the velocity time graph is equal to the displacement of the object

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a motion in where it increases constantly every second

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The area under the line on a velocity-time graph is equal to the displacement of the object. If the shape of the graph can be broken into simple graph.

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v = v 0 + a t

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khalifa

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This can be found in drag or straight line racing. Maybe even the straight dune climbing race.

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1a: 625m

1b: 122.5m

2a: 240miles

2b:15m

2c:0m

3: 378m

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This means that the object is getting faster at a fixed speed in each time interval such as increasing at a constant 5 meters each second.

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a -> Vf-Vi/t -> V=Vi+at

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4: 8 seconds

5: 15m/s^2

6: 24.992m

I am not sure about 6

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How long it took for the drag car to finish the entire strip and its speed upon the finish line.

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I learnt how to calculate the displacement using a Velocity-Time graph..

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understanding the equations of motion on page 9 and q6

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The link has expired.

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1a:625m

1b:122.5m

2a:240miles

2b:15m

2c:0m

3:378m

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this can be used in drag racing

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Drag race

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Displacement is given by the area under the velocity-time graph of the object and acceleration is given by the slope of velocity-time graph.

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Constant acceleration refers to motion where the speed increases by the same amount each second. The most notable and important example is free fall. When an object is thrown or dropped, it experiences a constant acceleration due to gravity, which has a constant value of approximately 10 meters per second squared.

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drag race

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by finding the area under the graph

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motion where the speed increases by the same amount each second.

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v=v0+at. v = v 0 + a t .

v0+v2=v0+12at. v 0 + v 2 = v 0 + 1 2 a t .

–v=v0+12at. v – = v 0 + 1 2 a t .

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by looking at the area under the graph

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find the area of the shape . the area under the line velocity- time graph is equal to the displacement of the object .

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1:625m

2:122.5

3a:240

3b:15

3c:90

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Q1,A. 25 x 25 = 625m

Q1,B. 1/2 x 7 x 35= 122.5m

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a golf ball

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how to calculate the displacment by getting the velocity and time

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a golf boll

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Q2,A. 40 x 6 = 240

Q2,B. 6 x 5 x 1/2 = 15

Q2,C. 15 x 12 = 180

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1: 625m

2: 122.5

3a: 240

3b: 15

3c:90

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To find the displacement we need to find the area under line graph.

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1:625m

2:122.5

3a:240

3c:90

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a ->vf-vi/t -> v=vi+at

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1a. area =bxh

1. area =25x25

area=625m

1b.area =1/2xbxh

area=1/2x7x35

area=122.5m

2a. area=bxh

area=6x40

area=240m

2b.area=1/2xbxh

area=1/2x5x6

area=15m

2.c area 1/2xbxh

area=1/2x12x15

area=90m

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Q1

-625m

-122.5m

Q2

-240 miles

-15m

-90m

Q3

-378

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1 A - 625m

1 B - 122.5m

2 A - 240 miles

2 B - 15m

2 C - 90m

3 - 378

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3

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a golf ball

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Q1A=625m

Q1B=122.5m

Q2A=240

Q2B=15

Q2C=90

Q3=70

Q3=308

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finding the area under the graph