changing motion
Post on 18-Jan-2016
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1. What is the shortest possible time in which a bacterium could drift at a velocity of 3.5 mm/s across a petri dish with a diameter of 8.4 cm ?
2. A child is pushing a shopping cart at a speed of 1.5 m/s. How long will it take this child to push the cart down an aisle with a length of 9.3 m ?
3. It takes you 9.5 minutes to walk with an average velocity of 1.2 m/s to the north from the bus stop to the museum entrance.
What is your displacement ?
Changing Motion
Acceleration
• Acceleration is any change in the velocity of an object with respect to time.
• The difference between two velocities divided by the time for the change.
• Acceleration has a magnitude and direction.
Acceleration
• It measures the rate of change in velocity.
• The SI unit for acceleration is m/s2
(vf –vi )• a = ---------- t
a = acceleration
vf = velocity final
vi = velocity initial t = time
Example #1
A car increases its velocity from 4 m/s to 25 m/s in 5.5 seconds. Find the car’s acceleration?
G: vi = 4 m/s vf = 25 m/s t = 5.5 s
U: a
(vf –vi )
E: a = ---------- t
S: substitute
S: 3.8 m/s2
Displacement with Constant Uniform Acceleration
• We know that the average velocity is equal to displacement divided by the time interval.
d
v avg = t
• For an object moving with constant acceleration, the average velocity is equal to the average of the initial velocity and the final velocity.
vf + vi
• v avg = 2
Displacement with Constant Uniform Acceleration
d (vf + vi)
t = v avg and v avg = 2
d (vf + vi)
t = v avg = 2
d (vf + vi)t = 2
d = ½ ( vf + vi ) t
Example #2
A car increases its velocity from 4 m/s to 25 m/s in
5.5 seconds. Find the car’s displacement while speeding up.
G: vi = 4 m/s vf = 25 m/s t = 5.5 s
U: d
E: d = ½ ( vf + vi ) t
S: substitute
S: 79.8 m
Example #3
G: vi = 25 m/s vf = 0 m/s d = 50 m
U: a ( vf – vi )E: a = t d = ½ ( vi + vf ) t
S: substitute
S: -6.3 m/s2
The car’s driver then applies the brakes and stops the car in 50 meters.
Find the car’s acceleration
A car increases its velocity from 4 m/s to 25 m/s in 5.5 seconds.
Example #4
A car increases its velocity from 4 m/s to 25 m/s in 5.5 seconds. The car’s driver then applies the brakes and stops the car in 50 meters.
Find the time to stop the car.
G:
U:
E:
S:
S: See work on example 3
A golf ball is dropped from a 10-meter tall building, strikes the ground at 12.5 m/s and rebounds at 5 m/s. Find the ball’s acceleration from the point it is dropped until it strikes the ground, if the time of impact equaled 0.025 seconds.
G: d = 10 m vi = 0 m/s vfdown = 12.5 m/s (down) vfup = 5 m/s (up) t = 0.025 s
U: a ( vf – vi )E: a = t
S: substitute
S: 500 m/s2
Example #5
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