Explanations and Derivations · Ima Hurryin is approaching a stoplight moving with a velocity of...
Transcript of Explanations and Derivations · Ima Hurryin is approaching a stoplight moving with a velocity of...
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Four Kinematic Equations
Explanations and Derivations
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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
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Motion Equation #1 Displacement with Constant Acceleration
…move t to the other side...
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Motion Equation #2 Velocity with Constant Acceleration
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Another Approach to Equation #2
y = mx + b
velo
city
(m/s
)
time (s)
vf = at + vi vf = vi + ator
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Motion Equation #3Displacement with Constant Acceleration
By substituting the expression for vf into our first equation for Dx, we get:
First equation:
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Motion Equation #4Final velocity after any displacement
See derivation in text book…
FINAL VELOCITY AFTER ANY DISPLACEMENT
vf2 = vi
2 + 2aDx
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Summary
Motion with Constant Acceleration
We now have all the equations we need to solve constant-acceleration problems.
vf2 = vi
2 + 2aDx
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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
t = 6.00 sec
vf = 0 m/s
Dx = ?
Choose an equation
Dx = 63.0 m
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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
Dx = 15.0 m
vi = 0 m/s
a = ?
Choose an equation
= 4.41 m/s2
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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
Dx = 15.0 m
vi = 0 m/s
a = 4.41 m/s2
Choose an equation
vf – vi = at
= 2.61 sec
t = ?
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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 = vi
2 + 2aDx
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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.
Dx = 56.3 m
vf2 = vi
2 + 2aDx
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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.
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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
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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).
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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
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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 = vi
2 + 2aDx
or
a = 2.86 m/s/st = 30. 9 s
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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 = vi
2 + 2aDx
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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
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• Due Mon:
Kinematic Equations Worksheet