Four Kinematic Equations
Explanations and Derivations
For an object moving with constant acceleration, the average velocity is
equal to the average of the initial velocity and the final velocity
We also know, of course, that
Motion Equation #1
Displacement with Constant Acceleration
…move t to the other side...
Motion Equation #2
Velocity with Constant Acceleration
velocity (m/s)
Another Approach to Equation #2
y = mx + b
vf = at + vi
time (s)
or
vf = vi + at
Motion Equation #3
Displacement with Constant Acceleration
By substituting the expression for vf into our first
equation for Dx, we get:
First equation:
Motion Equation #4
Final velocity after any displacement
See derivation in text book…
FINAL VELOCITY AFTER ANY DISPLACEMENT
vf2 = vi2 + 2aDx
Summary
Motion with Constant Acceleration
We now have all the equations we need to solve
constant-acceleration problems.
2
vf
=
2
vi
+ 2aDx
A car slows down uniformly from a speed of 21.0 m/s to
rest in 6.00 seconds. How far did it travel in that time?
The words “slowing down uniformly” implies that the car has a
constant acceleration.
vi = +21.0 m/s
Choose an equation
t = 6.00 sec
vf = 0 m/s
Dx = ?
Dx = 63.0 m
A world-class sprinter can burst out of the blocks and
reach essentially top speed (of about 11.5 m/s) in the first
15.0 m of the race. What is the average acceleration of this
sprinter, and how long does it take her to reach that speed?
vf = +11.5 m/s
Choose an equation
Dx = 15.0 m
vi = 0 m/s
a=?
= 4.41 m/s2
A world-class sprinter can burst out of the blocks and
reach essentially top speed (of about 11.5 m/s) in the first
15.0 m of the race. What is the average acceleration of this
sprinter, and
vf = +11.5 m/s
Choose an equation
Dx = 15.0 m
vi = 0 m/s
vf – vi = at
a = 4.41 m/s2
t=?
=
2.61 sec
Ima Hurryin is approaching a stoplight moving with a velocity of
+30.0 m/s. The light turns yellow, and Ima applies the brakes and
skids to a stop. If Ima accelerates at a rate of 8.00 m/s2, then
determine the displacement of the car during the skidding process.
(Note that the direction of the velocity and the acceleration vectors
are denoted by a + and a - sign.)
vf2 = vi2 + 2aDx
Ima Hurryin is approaching a stoplight moving with a velocity
of +30.0 m/s. The light turns yellow, and Ima applies the
brakes and skids to a stop. If Ima has an acceleration rate of
8.00 m/s2, determine the displacement of the car during the
skidding process.
vf2 = vi2 + 2aDx
Dx = 56.3 m
Ben Rushin is waiting at a stoplight. When it finally turns green,
Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10
seconds. Determine the displacement of Ben's car during this time
period.
Ben Rushin is waiting at a stoplight. When it finally turns green,
Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10
seconds. Determine the displacement of Ben's car during this time
period.
Dx = 50.4 m
A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the
skidding distance of the car (assume uniform acceleration).
A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the
skidding distance of the car (assume uniform acceleration).
Dx = 28.6 m
A plane has a takeoff speed of 88.3 m/s and requires 1365 m to
reach that speed. Determine the acceleration of the plane and
the time required to reach this speed.
vf2 = vi2 + 2aDx
or
a = 2.86 m/s/s
t = 30. 9 s
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.
vf2 = vi2 + 2aDx
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.
vi = 94.6 mi/hr
• Due Mon:
Kinematic Equations Worksheet