

Lesson 1: Describing Motion with WordsIntroduction to the Language of Kinematics 
Lesson 6: Kinematic Equations and ProblemSolvingKinematic Equations and GraphsLesson 4 of this unit at The Physics Classroom focused on the use of velocitytime graphs to describe the motion of objects. In that Lesson, it was emphasized that the slope of the line on a velocitytime graph is equal to the acceleration of the object and the area between the line and the time axis is equal to the displacement of the object. Thus, velocitytime graphs can be used to determine numerical values and relationships between the quantities displacement (d), velocity (v), acceleration (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 now two methods available to solve problems involving the numerical relationships between displacement, velocity, acceleration and time. In this part of Lesson 6, we will investigate the relationships between these two methods. Consider an object which moves with a constant velocity of +5 m/s for a time period of 5 seconds and then accelerates to a final velocity of +15 m/s over the next 5 seconds. Such a verbal description of motion can be represented by a velocitytime graph. The graph is shown below. The horizontal section of the graph depicts a constant velocity motion, consistent with the verbal description. The positively sloped (i.e., upward sloped) section of the graph depicts a positive acceleration, consistent with the verbal description of an object moving in the positive direction and speeding up from 5 m/s to 15 m/s. The slope of the line can be computed using the rise over run ratio. Between 5 and 10 seconds, the line rises from 5 m/s to 15 m/s and runs from 5 s to 10 s. This is a total rise of +10 m/s and a total run of 5 s. Thus, the slope (rise/run ratio) is (10 m/s)/(5 s) = 2 m/s^{2}. Using the velocitytime graph, the acceleration of the object is determined to be 2 m/s^{2} during the last five seconds of the object's motion. The displacement of the object can also be determined using the velocitytime graph. The area between the line on the graph and the timeaxis is representative of the displacement; this area assumes the shape of a trapezoid. As discussed in Lesson 4, the area of a trapezoid can be 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 area of the rectangle plus the area of the triangle. The calculation of these areas is shown below.
The total area (rectangle plus triangle) is equal to 75 m. Thus the displacement of the object is 75 meters during the 10 seconds of motion. The above discussion illustrates how a graphical representation of an object's motion can be used to extract numerical information about the object's acceleration and displacement. Once constructed, the velocitytime graph can be used to determine the velocity of the object at any given instant during the 10 seconds of motion. For example, the velocity of the object at 7 seconds can be determined by reading the ycoordinate value at the xcoordinate of 7 s. Thus, velocitytime graphs can be used to reveal (or determine) numerical values and relationships between the quantities displacement (d), velocity (v), acceleration (a) and time (t) for any given motion.
Now let's consider the same verbal description and the corresponding analysis using kinematic equations. The verbal description of the motion was: An object which moves with a constant velocity of +5 m/s for a time period of 5 seconds and then accelerates to a final velocity of +15 m/s over the next 5 seconds Kinematic equations can be applied to any motion for which the acceleration is constant. Since this motion has two separate acceleration stages, any kinematic analysis requires that the motion parameters for the first 5 seconds not be mixed with the motion parameters for the last 5 seconds. The table below lists the given motion parameters.
Note that the acceleration during the first 5 seconds is listed as 0 m/s^{2} despite the fact that it is not explicitly stated. The phrase constant velocity indicates a motion with a 0 acceleration. The acceleration of the object during the last 5 seconds can be calculated using the following kinematic equation. The substitution and algebra are shown here. 15 m/s  5 m/s = a*(5 s) 10 m/s = a*(5 s) (10 m/s)/(5 s) = a a = 2 m/s^{2} This value for the acceleration of the object during the time from 5 s to 10 s is consistent with the value determined from the slope of the line on the velocitytime graph. The displacement of the object during the entire 10 seconds can also be calculated using kinematic equations. Since this 10 seconds include two distinctly different acceleration intervals, the calculations for each interval must be done separately. This is shown below.
The total displacement during the first 10 seconds of motion is 75 meters, consistent with the value determined from the area under the line on the velocitytime graph. The analysis of this simple motion illustrates the value of these two representations of motion  velocitytime graph and kinematic equations. Each representation can be utilized to extract numerical information about unknown motion quantities for any given motion. The examples below provide useful opportunity for those requiring additional practice.

Given:

Find: 
(0 m/s)^{2} = (25.0 m/s)^{2} + 2 * (1.0 m/s^{2})*d
0.0 m^{2}/s^{2} = 625.0 m^{2}/s^{2} + (2.0 m/s^{2})*d
0.0 m^{2}/s^{2}  625.0 m^{2}/s^{2} = (2.0 m/s^{2})*d
(625.0 m^{2}/s^{2})/(2.0 m/s^{2}) =d
313 m = d
a. The velocitytime 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 axis. This area would be the area of the triangle plus the area of rectangle 1 plus the area of rectangle 2.
Area = 0.5*b_{tri}*h_{tri} + b_{rect1}*h_{rect1} + b_{rect2}*h_{rect2} Area = 0.5*(5.0 s)*(10.0 m/s) + (5.0 s)*(25.0 m/s) + (10.0 s)*(35.0 m/s)
Area = 25 m + 125 m + 350 m
Area = 500 m
b. The distance traveled can be calculated using a kinematic equation. The solution is shown here.
First find the d for the first 5 seconds:
Given:
v_{i} = 25.0 m/s t = 5.0 s
a = 2.0 m/s^{2}
Find:
d = ?? d = v_{i}*t + 0.5*a*t^{2} d = (25.0 m/s)*(5.0 s) + 0.5*(2.0 m/s^{2})*(5.0 s)^{2}
d = 125 m + 25.0 m
d = 150 m
Now find the d for the last 10 seconds:
Given:
v_{i} = 35.0 m/s t = 10.0 s
a = 0.0 m/s^{2}
Find:
d = ?? (Note: the velocity at the 5 second mark can be found from knowing that the car accelerates from 25.0 m/s at +2.0 m/s^{2} for 5 seconds. This results in a velocity change of a*t = 10 m/s, and thus a velocity of 35.0 m/s.) d = v_{i}*t + 0.5*a*t^{2}
d = (35.0 m/s)*(10.0 s) + 0.5*(0.0 m/s^{2})*(10.0 s)^{2}
d = 350 m + 0 m
d =350 m
The total distance for the 15 seconds of motion is the sum of these two distance calculations (150 m + 350 m):
distance = 500 m
a. The velocitytime graph for the motion is:b. The time to rise up and fall back down to the original height is twice the time to rise up to the peak. So the solution involves finding the time to rise up to the peak and then doubling it.
Given:
v_{i} = 40.0 m/s v_{f} = 0.0 m/s
a = 10.0 m/s^{2}
Find:
t_{up} = ?? 2*t_{up} = ??
v_{f} = v_{i} + a*t_{up} 0 m/s = 40 m/s + (10 m/s2)*t_{up}
(10 m/s^{2})*t_{up} = 40 m/s
t_{up} = (40 m/s)/(10 m/s^{2})
t_{up} = 4.0 s
2*t_{up} = 8.0 s
a. The velocitytime 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 axis. This area would be the area of the rectangle plus the area of the triangle.
Area = b_{rect}*h_{rect} + 0.5*b_{tri}*h_{tri} Area = (10.0 min)*(0.50 mi/min) + 0.5*(2.0 min)*(0.50 mi/min)
Area = 5 mi + 0.5 mi
Area = 5.5 mi
b. The distance traveled can be calculated using a kinematic equation. The solution is shown here.
First find the d for the first 10 minutes:
Given:
v_{i} = 0.50 mi/min t = 10.0 min
a = 0.0 mi/min^{2}
Find:
d = ?? d = v_{i}*t + 0.5*a*t^{2} d = (0.50 mi/min)*(10.0 min) + 0.5*(0.0 mi/min^{2})*(10.0 min)^{2}
d = 5.0 mi + 0 mi
d = 5.0 mi
Now find the d for the last 2 minutes:
Given:
v_{i} = 0.50 mi/min t = 2.0 min
a = 0.25 mi/min^{2}
Find:
d = ?? d = v_{i}*t + 0.5*a*t^{2} d = (0.50 mi/min)*(2.0 min) + 0.5*(0.25 m/s^{2})*(2.0 min)^{2}
d = 1.0 mi + (0.5 mi)
d = 0.5 mi
The total distance for the 12 minutes of motion is the sum of these two distance calculations (5.0 mi + 0.5 mi):
distance = 5.5 mi
a. The velocitytime 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 axis.
Area = 0.5*b*h = 0.5*(4.5 s)*(45.0 m/s) Area = 101 m
b.
Given:
v_{i} = 45.0 m/s v_{f} = 0.0 m/s
a = 10.0 m/s^{2}
Find:
d = ?? v_{f}^{2} = v_{i}^{2} + 2*a*d (0 m/s)^{2} = (45.0 m/s)^{2} + 2 * (10.0 m/s^{2})*d
0.0 m^{2}/s^{2} = 2025.0 m^{2}/s^{2} + (20.0 m/s^{2})*d
0.0 m^{2}/s^{2}  2025.0 m^{2}/s^{2} = (20.0 m/s^{2})*d
(2025.0 m^{2}/s^{2})/(20.0 m/s^{2}) =d
101 m =d
Since the accident pileup is less than 101 m from Vera, she will indeed hit the pileup before completely stopping (unless she veers aside).
a. The velocitytime 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 axis. This area would be the area of the triangle plus the area of rectangle 1 plus the area of rectangle 2.
Area = 0.5*b_{tri}*h_{tri} + b_{1}*h_{1} + b_{2}*h_{2} Area = 0.5*(5.0 s)*(15.0 m/s) + (10.0 s)*(30.0 m/s) + (5.0 s)*(30.0 m/s)
Area = 37.5 m + 300 m + 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:
v_{i} = 30.0 m/s t = 10.0 s
a = 0.0 m/s^{2}
Find:
d = ?? d = v_{i}*t + 0.5*a*t^{2} d = (30.0 m/s)*(10.0 s) + 0.5*(0.0 m/s^{2})*(10.0 s)^{2}
d = 300 m + 0 m
d =300 m
Now find the d for the last 5 seconds:
Given:
v_{i} = 30.0 m/s t = 5.0 s
a = 3.0 m/s^{2}
Find:
d = ?? d = v_{i}*t + 0.5*a*t^{2}
d = (30.0 m/s)*(5.0 s) + 0.5*(3.0 m/s^{2})*(5.0 s)^{2}
d = 150 m + 37.5 m
d = 187.5 m
The total distance for the 15 seconds of motion is the sum of these two distance calculations (300 m + 187.5 m):
distance = 488 m
Lesson 6: Kinematic Equations and ProblemSolving
 The Kinematic Equations
 Kinematic Equations and ProblemSolving
 Kinematic Equations and Free Fall
 Sample Problems and Solutions
 Kinematic Equations and Graphs
Other Resources: Physics Home Page  Multimedia Physics Studios  Shockwave Physics Studios 
Minds On Physics Internet Modules  The Review Session
© Tom Henderson
19962007