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1.A water skier is pulled by a boat with an initial velocity of 10 m/s, The boat then accelerates to a velocity of 15 m/s over the next 12.5 s. What is.

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Presentation on theme: "1.A water skier is pulled by a boat with an initial velocity of 10 m/s, The boat then accelerates to a velocity of 15 m/s over the next 12.5 s. What is."— Presentation transcript:

1 1.A water skier is pulled by a boat with an initial velocity of 10 m/s, The boat then accelerates to a velocity of 15 m/s over the next 12.5 s. What is the final velocity of the water skier? 2.A rocket is launched from Earth and reaches a velocity of 500 m/s. If it has a constant acceleration of 12 m/s 2, how long will it take to reach this velocity? 3.A ball is dropped from a height of 20 m. If it accelerates at 9.80 m/s 2 and has a final velocity of 50 m/s, how long will it take for the ball to hit the ground? 4.A person standing on the Golden Gate Bridge, drops a rock over the edge to the water. If the rock falls for 13.68 s and has a velocity of 134.1 m/s just before hitting the water, what is the rock’s acceleration ?

2 Freefall

3 Freefall Objectives 1. Qualitatively and quantitatively describe the motion (i.e., position, velocity and acceleration) of an object in freefall. 2. Solve freefall problems.

4 Freefall Galileo found that in the absence of air resistance all objects dropped near the surface of a planet will fall with the same acceleration. Freely falling bodies undergo constant acceleration. For the purpose of the calculations at this time, we will neglect air resistance.

5 Acceleration Due to Gravity The constant of gravity on Earth is g = 9.8 m/s 2 The gravity constant on other planets will differ and can be calculated. In our calculations acceleration (a) is equal to gravity (g).

6 What Goes Up Must Come Down We know that when we throw an object up into the air, it will continue to move upward for some time, stop momentarily at the peak, and then change direction and begin to fall. Because the object changes direction, it may seem like the velocity and acceleration are both changing, but in reality, objects thrown into the air have a downward acceleration as soon as they are released.

7 Formulas d = v i t + ½at 2 v f = √ v i 2 + 2ad or v f 2 = v i 2 + 2ad v f = v i + at d = ½ (v f + v i )t

8 Example #1 The Demon Drop ride at Cedar Point Amusement Park falls freely for 1.5 s after starting from rest. What is its velocity at the end of the 1.5 s? G: t = 1.5 s v i = 0 m/s a = g = -9.80 m/s 2 U:vfU:vf E:v f = v i + at S: v f = (0 m/s) + (-9.8 m/s 2 )(1.5 s) S: - 14.70 m/s

9 Example #2 The Demon Drop ride at Cedar Point Amusement Park falls freely for 1.5 s after starting from rest. How Far did it fall during 1.5 s? G: t = 1.5 s v i = 0 m/s a = g = -9.80 m/s 2 v f = -14.70 m/s (information from example 1) U:dU:d E:d = v i t + 1/2at 2 S: d = (0 m/s)(1.5 s) + (1/2)(-9.8 m/s 2 )(1.5 s) 2 S: - 11.03 m

10 Example #3 A tennis ball is thrown straight up with an initial velocity of 22.5 m/s. If it is caught at the same distance above the ground, how high did the ball rise ? G::v i = 22.5 m/s a = g = -9.80 m/s 2 v f = 0 m/s U: d E:v f 2 = v i 2 +2ad S: d = ( (0 m/s) 2 – (22.5 m/s) 2 ) / ( (2)(-9.8 m/s 2 ) ) S: 25.83 m d = (v f 2 - v i 2 ) / (2a)→

11 Example #4 A tennis ball is thrown straight up with an initial velocity of 22.5 m/s. If it is caught at the same distance above the ground, how long does the ball stay in the air ? G::v i = 22.5 m/s a = g = -9.80 m/s 2 v f = 0 m/s d = 25.83 m (information from previous problem) U: t E:v f = v i + at S: t = ( (0 m/s) – (22.5 m/s) ) / (-9.8 m/s 2 ) S: 2.30 s t = (v f – v i ) / a→

12 Example #5 A spaceship far from any planet or star and is not affected by any gravity accelerates up from 65.0 m/s to 162.0 m/s in 10.0 s. How far does it travel during the acceleration ? G::v i = 65.0 m/s v f = 162.0 m/s t = 10.0 s U:dU:d E:d = (1/2)(v f + v i )t S: d = (1/2)(162.0 m/s + 65 m/s)(10.0 s) S: 1.135 x 10 3 m

13 Example #6 An express train traveling at 36.0 m/s, is accidentally sidetracked onto a local train track. The express train engineer spots a local train 1.00 x 10 3 m ahead on the same track and traveling in the same direction. The local engineer is unaware of the situation. The express engineer jams on the brakes and slows the express train at a constant rate of 3.00 m/s 2. If the speed of the local train is a constant 11.00 m/s, will the express train be able to stop or will there be a collision ? In the 12.00 s it take for the express train to stop the local train will travel 232.00 m and the express train will travel 216.00 m. There will be no collision.

14 Freefall Your mission, should you choose to accept it. Complete the On-Line Homework assignment #1.


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