1 / 11

Changing Motion

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 ?.

quanda
Download Presentation

Changing Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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 ?

  2. Changing Motion Changing Motion

  3. 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.

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. Example #3 A car increases its velocity from 4 m/s to 25 m/s in 5.5 seconds. 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

  10. 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

  11. Example #5 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

More Related