Lesson iv of this unit at The Physics Classroom focused on the utilize of velocity-time graphs to draw the motion of objects. In that Lesson, information technology was emphasized that the slope of the line on a velocity-time graph is equal to the acceleration of the object and the expanse between the line and the fourth dimension axis is equal to the displacement of the object. Thus, velocity-time graphs tin be used to decide numerical values and relationships between the quantities displacement (d), velocity (v), dispatch (a) and time (t). In Lesson 6, the focus has been upon the use of four kinematic equations to describe the motion of objects and to predict the numerical values of one of the four motion parameters - displacement (d), velocity (v), acceleration (a) and time (t). Thus, there are at present ii methods bachelor to solve problems involving the numerical relationships betwixt deportation, velocity, acceleration and time. In this part of Lesson 6, we will investigate the relationships betwixt these ii methods.
Instance Problem - Graphical Solution
Consider an object that moves with a constant velocity of +5 grand/due south for a time flow of five seconds and then accelerates to a concluding velocity of +xv m/due south over the adjacent 5 seconds. Such a verbal description of motility can be represented by a velocity-time graph. The graph is shown beneath.
The horizontal section of the graph depicts a constant velocity movement, consistent with the verbal description. The positively sloped (i.e., upward sloped) section of the graph depicts a positive acceleration, consistent with the verbal clarification of an object moving in the positive direction and speeding up from five yard/southward to 15 grand/s. The slope of the line tin can be computed using the rise over run ratio. Between five and x seconds, the line rises from 5 grand/due south to 15 one thousand/southward and runs from 5 due south to 10 southward. This is a full rise of +10 m/s and a full run of 5 s. Thus, the gradient (rise/run ratio) is (10 m/s)/(v s) = 2 m/s2. Using the velocity-fourth dimension graph, the acceleration of the object is adamant to exist 2 m/s2 during the last five seconds of the object's motion. The deportation of the object can also be adamant using the velocity-fourth dimension graph. The area betwixt the line on the graph and the time-axis is representative of the deportation; this area assumes the shape of a trapezoid. As discussed in Lesson 4, the area of a trapezoid tin can exist equated to the area of a triangle lying on top of the area of a rectangle. This is illustrated in the diagram below.
The total area is then the surface area of the rectangle plus the area of the triangle. The adding of these areas is shown beneath.
| Rectangle | Triangle |
Area = base of operations * acme
Area = (10 s) * (5 g/s) Area = 50 chiliad | Area = 0.v * base * meridian
Area = 0.v * (five s) * (x m/due south) Surface area = 25 m |
The total area (rectangle plus triangle) is equal to 75 1000. Thus the displacement of the object is 75 meters during the x seconds of movement.
The above discussion illustrates how a graphical representation of an object's move can be used to extract numerical information about the object'due south dispatch and displacement. Once synthetic, the velocity-time graph can be used to determine the velocity of the object at any given instant during the x seconds of motion. For example, the velocity of the object at 7 seconds tin can be determined by reading the y-coordinate value at the ten-coordinate of 7 due south. Thus, velocity-time graphs tin can be used to reveal (or determine) numerical values and relationships between the quantities displacement (d), velocity (v), acceleration (a) and time (t) for whatsoever given motion.
Example Problem - Solution Using Kinematic Equation
Now let'due south consider the aforementioned verbal description and the corresponding analysis using kinematic equations. The exact description of the motion was:
An object that moves with a constant velocity of +five thousand/s for a time period of 5 seconds and then accelerates to a terminal velocity of +15 one thousand/due south over the side by side five seconds
Kinematic equations tin can be applied to any movement for which the dispatch is constant. Since this motion has two separate acceleration stages, any kinematic analysis requires that the motion parameters for the outset 5 seconds not exist mixed with the move parameters for the last five seconds. The table below lists the given motion parameters.
| t = 0 s - five southward | t = 5 s - 10 s |
vi = 5 one thousand/s
vf = five m/south t = five s a = 0 m/due south2 | vi = 5 m/s
vf = 15 m/s t = v s |
Note that the acceleration during the kickoff 5 seconds is listed as 0 one thousand/stwo despite the fact that it is not explicitly stated. The phrase constant velocity indicates a motion with a 0 acceleration. The dispatch of the object during the last v seconds tin can be calculated using the post-obit kinematic equation.
5f = vi + a*t
The substitution and algebra are shown here.
15 m/s = 5 k/s + a*(5 due south)
15 chiliad/s - 5 1000/south = a*(v southward)
10 m/south = a*(5 s)
(10 k/southward)/(5 s) = a
a = 2 1000/s2
This value for the dispatch of the object during the time from 5 s to 10 south is consistent with the value determined from the gradient of the line on the velocity-time graph.
The displacement of the object during the entire 10 seconds tin can also be calculated using kinematic equations. Since these 10 seconds include ii distinctly different dispatch intervals, the calculations for each interval must be washed separately. This is shown below.
| t = 0 s - 5 s | t = 5 s - 10 s |
d = fivei*t + 0.five*a*tii
d = (5 one thousand/s)*(5 s) +0.five*(0 m/southwardtwo)*(v s)2
d = 25 k + 0 m
d = 25 g | d = ((vi + vf)/2)*t
d = ((v m/s + 15 m/s)/2)*(5 due south)
d = (10 m/s)*(5 s)
d = l m |
The total deportation during the first 10 seconds of motion is 75 meters, consistent with the value adamant from the area under the line on the velocity-time graph.
The assay of this simple motion illustrates the value of these two representations of motion - velocity-time graph and kinematic equations. Each representation tin can be utilized to excerpt numerical data virtually unknown move quantities for any given motion. The examples below provide useful opportunity for those requiring additional practice.
Check Your Understanding
1.Rennata Gas is driving through town at 25.0 m/s and begins to accelerate at a abiding charge per unit of -i.0 m/south2. Eventually Rennata comes to a consummate stop. a. Represent Rennata's accelerated movement past sketching a velocity-time graph. Use the velocity-time graph to determine this distance.
b. Use kinematic equations to calculate the distance that Rennata travels while decelerating.
2. Otto Emissions is driving his machine at 25.0 k/s. Otto accelerates at two.0 yard/s2 for 5 seconds. Otto then maintains a constant velocity for x.0 more than seconds. a. Stand for the 15 seconds of Otto Emission's motion by sketching a velocity-time graph. Use the graph to decide the altitude that Otto traveled during the unabridged xv seconds.
b. Fin marry, break the move into its 2 segments and use kinematic equations to calculate the full distance traveled during the unabridged 15 seconds.
3. Luke Autbeloe, a human cannonball artist, is shot off the border of a cliff with an initial upward velocity of +40.0 yard/s. Luke accelerates with a abiding downward acceleration of -10.0 m/stwo (an approximate value of the acceleration of gravity). a. Sketch a velocity-fourth dimension graph for the first eight seconds of Luke'southward move.
b. Use kinematic equations to make up one's mind the time required for Luke Autbeloe to drop back to the original peak of the cliff. Indicate this fourth dimension on the graph.
4. Chuck Carriage travels with a abiding velocity of 0.5 mile/minute for x minutes. Chuck then decelerates at -.25 mile/min2 for two minutes. a. Sketch a velocity-tim e graph for Chuck Railroad vehicle's move. Use the velocity-time graph to determine the total distance traveled past Chuck Carriage during the 12 minutes of motion.
b. F inally, pause the movement into its two segments and employ kinematic equations to determine the total distance traveled by Chuck Railroad vehicle.
5. Vera Side is speeding downward the interstate at 45.0 m/southward. Vera looks ahead and observes an accident that results in a pileup in the centre of the route. Past the fourth dimension Vera slams on the breaks, she is 50.0 chiliad from the pileup. She slows downwardly at a charge per unit of -10.0 yard/sii. a. Construct a velocity-time plot for Vera Side'due south motility. Use the plot to determine the distance that Vera would travel prior to reaching a complete stop (if she did not collide with the pileup).
b. Utilize kinematic equations to determine the distance that Vera Side would travel prior to reaching a complete finish (if she did not collide with the pileup). Will Vera hit the cars in the pileup? That is, will Vera travel more than 50.0 meters?
six. Earl Eastward. Bird travels 30.0 m/s for 10.0 seconds. He then accelerates at three.00 chiliad/s2 for 5.00 seconds. a.Construct a velocity-time graph for Earl E. Bird's motion. Apply the plot to make up one's mind the total distance traveled.
b. Divide the motility of the Earl E. Bird into the ii time segments and utilise kinematic equations to calculate the total displacement.
Solutions to To a higher place Questions
Solution to Question 1
a. The velocity-time graph for the motion is:
The distance traveled can be found by a calculation of the area between the line on the graph and the time centrality.
Area = 0.five*b*h = 0.5*(25.0 s)*(25.0 thousand/s)
Area = 313 g
b. The distance traveled tin be calculated using a kinematic equation. The solution is shown hither.
| Given: | vi = 25.0 m/s | vf = 0.0 g/s | a = -1.0 1000/stwo | | Observe: d = ?? |
vf 2 = vi 2 + ii*a*d
(0 k/due south)2 = (25.0 one thousand/southward)2 + 2 * (-1.0 k/s2)*d
0.0 m2/s2 = 625.0 m2/sii + (-ii.0 m/southward2)*d
0.0 one thousand2/southwardtwo - 625.0 m2/s2 = (-2.0 chiliad/s2)*d
(-625.0 m2/southward2)/(-2.0 m/south2) = d
313 m = d
Solution to Question 2
a. The velocity-time graph for the motility is:
The distance traveled tin can be found by a calculation of the area between the line on the graph and the fourth dimension axis. This area would exist the surface area of the triangle plus the area of rectangle i plus the area of rectangle 2.
Expanse = 0.five*btri*htri + brect1*hrect1 + brect2*hrect2
Expanse = 0.5*(v.0 s)*(10.0 m/s) + (5.0 s)*(25.0 1000/south) + (10.0 s)*(35.0 m/s)
Area = 25 m + 125 m + 350 m
Area = 500 1000
b. The distance traveled tin can exist calculated using a kinematic equation. The solution is shown here.
First find the d for the first 5 seconds:
| Given: | vi = 25.0 yard/s | t = 5.0 s | a = 2.0 thou/southward2 | | Find: d = ?? |
d = (25.0 m/due south)*(5.0 south) + 0.5*(2.0 m/due south2)*(5.0 s)2
d = 125 m + 25.0 m
d = 150 thou
At present find the d for the last x seconds:
| Given: | vi = 35.0 m/south | t = x.0 south | a = 0.0 chiliad/s2 | | Notice: d = ?? |
(Note: the velocity at the v 2nd mark can be constitute from knowing that the car accelerates from 25.0 grand/due south at +ii.0 thou/due south2 for 5 seconds. This results in a velocity change of a*t = 10 m/s, and thus a velocity of 35.0 1000/s.)
d = vi*t + 0.5*a*t2
d = (35.0 chiliad/due south)*(10.0 s) + 0.5*(0.0 one thousand/southii)*(10.0 southward)2
d = 350 thousand + 0 m
d =350 thousand
The total altitude for the xv seconds of motility is the sum of these 2 distance calculations (150 m + 350 g):
distance = 500 k
Solution to Question 3
a. The velocity-time graph for the motility is:
b. The time to rise up and fall back down to the original acme is twice the time to rise up to the peak. So the solution involves finding the time to rise upwards to the peak and and then doubling information technology.
| Given: | vi = 40.0 m/s | vf = 0.0 thou/due south | a = -10.0 m/s2 | | Find: tupwards = ?? 2*tupwards = ?? |
vf = fivei + a*tupwardly
0 g/south = forty m/s + (-ten m/s2)*tup
(ten m/s2)*tup = 40 thousand/s
tup = (40 yard/s)/(x m/s2)
tup = 4.0 southward
2*tup = eight.0 s
Solution to Question iv
a. The velocity-time graph for the motion is:
The distance traveled can exist plant by a calculation of the expanse between the line on the graph and the time axis. This area would be the expanse of the rectangle plus the area of the triangle.
Area = brect*hrect + 0.v*btri*htri
Expanse = (10.0 min)*(0.50 mi/min) + 0.5*(2.0 min)*(0.fifty mi/min)
Area = 5 mi + 0.v mi
Area = five.five mi
b. The distance traveled tin can be calculated using a kinematic equation. The solution is shown here.
First observe the d for the beginning x minutes:
| Given: | fivei = 0.fifty mi/min | t = x.0 min | a = 0.0 mi/min2 | | Find: d = ?? |
d = (0.50 mi/min)*(ten.0 min) + 0.5*(0.0 mi/mintwo)*(10.0 min)2
d = 5.0 mi + 0 mi
d = 5.0 mi
Now find the d for the last 2 minutes:
| Given: | 5i = 0.50 mi/min | t = two.0 min | a = -0.25 mi/min2 | | Find: d = ?? |
d = vi*t + 0.5*a*t2
d = (0.50 mi/min)*(2.0 min) + 0.5*(-0.25 m/southward2)*(ii.0 min)2
d = 1.0 mi + (-0.v mi)
d = 0.5 mi
The total distance for the 12 minutes of motion is the sum of these ii distance calculations (5.0 mi + 0.5 mi):
distance = five.5 mi
Solution to Question v
a. The velocity-time graph for the motion is:
The distance traveled tin can exist found by a calculation of the surface area betwixt the line on the graph and the time centrality.
Surface area = 0.five*b*h = 0.v*(four.5 s)*(45.0 m/due south)
Surface area = 101 chiliad
b.
| Given: | vi = 45.0 m/s | 5f = 0.0 thousand/s | a = -10.0 k/sii | | Find: d = ?? |
fivef 2 = fivei two + two*a*d
(0 m/due south)ii = (45.0 m/south)2 + 2 * (-10.0 grand/s2)*d
0.0 mii/southward2 = 2025.0 mii/stwo + (-20.0 1000/due south2)*d
0.0 g2/s2 - 2025.0 yard2/due southii = (-twenty.0 m/stwo)*d
(-2025.0 thousandii/southward2)/(-xx.0 one thousand/s2) =d
101 m =d
Since the accident pileup is less than 101 m from Vera, she will indeed hitting the pileup before completely stopping (unless she veers bated).
Solution to Question 6
a. The velocity-time graph for the motion is:
The distance traveled tin be constitute by a calculation of the area between the line on the graph and the fourth dimension axis. This area would be the surface area of the triangle plus the area of rectangle 1 plus the area of rectangle ii.
Expanse = 0.v*btri*htri + bi*hi + btwo*htwo
Surface area = 0.5*(v.0 s)*(15.0 m/southward) + (10.0 south)*(30.0 m/south) + (5.0 s)*(30.0 m/s)
Area = 37.5 m + 300 yard + 150 m
Area = 488 m
b. The distance traveled can be calculated using a kinematic equation. The solution is shown here.
First find the d for the first 10 seconds:
| Given: | vi = xxx.0 m/south | t = 10.0 southward | a = 0.0 yard/s2 | | Find: d = ?? |
d = 5i*t + 0.5*a*t2
d = (30.0 yard/s)*(10.0 due south) + 0.v*(0.0 g/s2)*(x.0 s)2
d = 300 m + 0 m
d =300 m
Now detect the d for the last 5 seconds:
| Given: | vi = 30.0 m/southward | t = 5.0 s | a = 3.0 m/southward2 | | Detect: d = ?? |
d = vi*t + 0.5*a*t2
d = (thirty.0 yard/s)*(5.0 s) + 0.five*(three.0 m/s2)*(5.0 s)ii
d = 150 m + 37.5 one thousand
d = 187.v thousand
The total altitude for the 15 seconds of movement is the sum of these two altitude calculations (300 m + 187.5 m):
distance = 488 thousand
