Name_____________________________ Per___________ Date_________
Creating the Kinematic Equations
Objective: We want to develop equations that will tell us the position of an object at any time, or the speed of
an object any time, if it is accelerating. To do that, we need to derive the equations, from the ones we know.
1. What is the equation for velocity (v)?
V=Δx/t
2. What is the equation for acceleration (a)?
a=Δv/t
3. What is the equation for displacement (Δx)?
Δx=xf-xi
4. What is the equation for change in velocity (Δv)?
Δv=vf-vi
5. Final position (constant velocity)
a. Using your equations for v and Δx, create an equation for xf, in terms of v, t, and xi.
Xf= vt + xi
b. You now have an equation that can be used to solve for xf, if you know the other variables.
c. For an object moving a constant velocity, the graph of position versus time looks like (HORIZONTAL
LINE, LINEAR LINE WITH SLOPE, CURVED LINE)
d. For an object moving with constant velocity, the position (DOES NOT CHANGE, CHANGES AT A
CONSTANT RATE, CHANGES AT A DIFFERENT RATE AS TIME INCREASES)
e. Write the math equation for a line.
Y=mx + b
f. Compare the two equations. The equations should have the same arrangement, but the letters will be
different. To what do the letters correspond?
Y=xf m= v x= t b= xi
g. A skater is moving at 12 m/s when he reaches the 50 m mark in a race. If he continues to skate at 12 m/s
for 7 seconds, what will his final position be?
Xf= vt + xi = (12 m/s)(7s)+50 = 134 m
Sign off___________
6. Final velocity (constant acceleration)
a. Using your equations a and Δv, create an equation for vf, in terms of a, t, and vi.
Vf=Vi + at
b. You now have an equation that can be used to solve for vf, if you know the other variables.
c. For an object moving a constant acceleration, the graph of velocity versus time looks like (HORIZONTAL
LINE, LINEAR LINE WITH SLOPE, CURVED LINE)
d. For an object moving with constant acceleration, the velocity (DOES NOT CHANGE, CHANGES AT A
CONSTANT RATE, CHANGES AT A DIFFERENT RATE AS TIME INCREASES)
e. Write the math equation for a line.
Y= mx +b
f. Compare the two equations. The equations should have the same arrangement, but the letters will be
different. To what do the letters correspond?
Y= Vf m= a x= t b= Vi
g. A cyclist is riding at 4 m/s when he reaches the top of a hill. If he accelerates down the hill at 1.2 m/s/s
for 8 seconds, what will his final velocity be?
Vf=Vi + at = Vf = 4 m/s +(1.2)(8) = 13.6 m/s
7. Final position (constant acceleration)
a. For an object moving a constant acceleration, the graph of position versus time looks like
(HORIZONTAL LINE, LINEAR LINE WITH SLOPE, QUADRATIC)
b. For an object moving with constant acceleration, the position (DOES NOT CHANGE, CHANGES AT A
CONSTANT RATE, CHANGES AT A DIFFERENT RATE AS TIME INCREASES)
c. If an object is accelerating, can we use the equation xf = vt + xi to determine the object’s final position?
Why? No, because that equation is for constant velocity, and if the object is accelerating, the velocity will
be changing.
d. Since the speed of an object changes, the amount of distance traveled each time interval is different. And
we can’t use the equation of a line to represent the motion. So we need to find an equation that gives us a
curve, specifically a quadratic.
b b 2 4ac
e. Write the math equation for a quadratic. Not the quadratic equation ( x )
2a
Y= Ax2+Bx+C
2
f. To determine the final position of an object we use the equation xf = 1/2at + vit + xi.
g. Compare the two equations. The equations should have the same arrangement, but the letters will be
different. To what do the letters correspond?
Y= Xf A=1/2a B= Vi C= Xi x= t
h. Solve for xf when a=4 m/s/s, t=5 sec, vi=10 m/s, and xi=20m.
xf = 1/2at2 + vit + xi. Xf = ½(4)(5)2+(10)(5)+20 = 120 m
Sign off___________
i. Using the equation for xf, what happens to the final position if we keep everything constant, but increase
the initial speed?
Xf will increase
1. Try it! Solve for xf when a=4 m/s/s, t=5 sec, vi=20 m/s, and xi=20m.
xf = 1/2at2 + vit + xi. Xf = ½(4)(5)2+(20)(5)+20 = 170 m
2. How does this compare to your original answer (part h)? (MORE THAN, LESS THAN,
EQUAL)
j. Using the equation for xf, what happens to the final position if we keep everything constant, but decrease
the initial position?
Xf will decrease
1. Try it! Solve for xf when a=4 m/s/s, t=5 sec, vi=10 m/s, and xi=5m.
xf = 1/2at2 + vit + xi. Xf = ½(4)(5)2+(10)(5)+5 = 105 m
2. How does this compare to your original answer (part h)? (MORE THAN, LESS THAN,
EQUAL)
k. Using the equation for xf, what happens to the final position if we keep everything constant, but increase
the acceleration?
Xf will increase
1. Try it! Solve for xf when a=12 m/s/s, t=5 sec, vi=10 m/s, and xi=20m.
xf = 1/2at2 + vit + xi. Xf = ½(12)(5)2+(10)(5)+5 = 220 m
2. How does this compare to your original answer (part h)? (MORE THAN, LESS THAN,
EQUAL)
l. A racing car is moving at125 ft/s when it comes around a sharp turn. The car accelerates at 15 ft/s/s for
10 seconds. Assuming the initial position of the car is 0, how many feet does the car travel in 10 seconds?
xf = 1/2at2 + vit + xi. Xf = ½(15 ft/s/s)(10)2+(125ft/s)(10)+0 = 2000 ft
m. Simplify the equation for xf, when the xi = zero.
xf = 1/2at2 + vit
n. Simplify the equation for xf, when both xi = zero and vi = zero.
xf = 1/2at2
Sign off___________