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Warm Up Question

Learn about uniformly accelerated motion and how to solve related problems using the four equations of motion. Includes step-by-step problem solving strategies.

jstacey
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Warm Up Question

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  1. Warm Up Question • 1. A tricycle racer passes the 2.00 meter line at 2.00 seconds. She then passes the 10.0 meter line at 10.00 seconds. Calculate her average speed. • 2. What would we need to know to express her average velocity? • 3. Knowing that she is behind her freshman competitor (who is a better tricyclist), she accelerates from 0.750 m/s to 1.25 m/s in 2.00 seconds. What is her rate of acceleration?

  2. Uniformly Accelerated Motion Physics Homework: See Unit Outline

  3. Notation Change: A different notation will be used to show initial and final values of some variables than were used yesterday. vo will be used for the initial velocity, rather than vi , or v1 v will be used for the final velocity rather than vf. , or v2 Dx = x – xo is the displacement, the distance between the endpoints of the motion of the object. Dt = t – to is the elapsed time. Any variable with a subscript ‘o’ is assumed an initial value of that quantity. If the variable does not have the subscript ‘o’, than it is assumed to be a later value, such as the value at the end of the time interval.

  4. I. Introduction In general, the motion of an object may be very complicated. The position, velocity, and acceleration may change over time. Today we will look at a special case, uniformly accelerated motion. acceleration The term “uniformly accelerated” means that the ________________ of the object remains _____________ over time. constant average If the acceleration stays constant over time, then the _____________ acceleration always equals the _______________ acceleration. instantaneous In other words, ‘a’ is the instantaneous acceleration and ‘aave’ is the average acceleration.

  5. initial The time is always measured as an interval, from an ____________ time of to to a __________ time of t. final To simplify the equations, we will always take the initial time to be __________ seconds. zero final elapsed With this change, the ______ time always equals the ________ time.

  6. The equations of motion introduced yesterday can be written as: Average velocity Average acceleration

  7. These equations will be used to derive 4 new equations. These equations only apply if the acceleration remains ___________ !! constant You are not required to memorize these derivations. The equations will be given to you on the exam. The first equation is just a rearrangement of the average acceleration equation. Since the acceleration is assumed constant, the variable representing instantaneous acceleration will be used exclusively. From rearrange to get and finally

  8. The second equation is just a restatement of the average velocity. The third equation will require manipulating and combining these first two equations. Solve the second equation for v, and substitute into the first equation.

  9. Thus: Rearrange to solve for Dx: Finally:

  10. The fourth equation will also require manipulating and combining the first two equations. Solve the first equation for t, and substitute into the second equation. Simplify to get:

  11. Cross multiply the terms: Simplify and rearrange:

  12. The four equations can be summarized on the following table. Also shown on the table are the 5 variables used in the equations. Each equation contains only 4 of the 5 variables. By knowing which variables are in each equation, you can decide which equation to use to solve a given problem. √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √

  13. Problem Solving Strategies: There are a few basic rules that one can follow to solve physics word problems. Every problem is different, but these guidelines are general enough to apply to all situations. 1. Read the problem. 2. Reread the problem. 3. Draw a picture. 4. Write down what is given in the problem, label on the drawing also. 5. Write down what is unknown. What are you being asked to find? 6. Decide which equation or equations best suit the given information. 7. Solve the problem.

  14. Example 1: A car accelerates from 10.0 m/s to 30.0 m/s in a time of 2.50 seconds. (a) What is the acceleration of the car? Given information: vo = 10.0 m/s t = 2.50 s v = 30.0 m/s vo Unknown: a = ? Equation containing v, vo, t, and a? Solve for a: Finally:

  15. (b) What is the distance the car travels? Given information: vo = 10.0 m/s t = 2.50 s v = 30.0 m/s a = 8.00 m/s2 Unknown information: Dx = ? Which equation? First option:

  16. Second option:

  17. Example 2: A duck accelerates from 10.0 cm/s to 30.0 cm/s in a distance of 25.0 cm. What is the acceleration of the duck? Given information: vo = 10.0 cm/s = 0.100 m/s Dx = 25.0 cm = 0.250 m v = 30.0 cm/s = 0.300 m/s Unknown: a = ? Equation containing v, vo, Dx, and a? Solve for a:

  18. Example 3: A motorcycle is moving at 30.0 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the first 3.00 s the brakes are applied, the motorcycle slows to 15.0 m/s. • Find the deceleration. • What distance does the motorcycle travel from the instant braking begins until it comes to a complete rest? Given information: vo = 30.0 m/s t = 3.00 s v = 15.0 m/s Unknown information: a = ? Equation containing v, vo, t, and a? Solve for a:

  19. Finally: The (-) sign means the acceleration points opposite to the direction of the velocity of the object. Now use the acceleration to solve for the distance traveled. Given information: vo = 30.0 m/s a = - 5.00 m/s2 v = 0 = rest or complete stop Unknown information: Dx = ? Equation containing v, vo, Dx, and a? Solve for Dx:

  20. Example 4: Two trains, one traveling at 72 km/h and the other traveling at 144 km/h, are headed towards one another on straight, level track. When the trains are 950 m apart, each engineer sees the other’s train and applies the brakes. The brakes slow each train at a rate of 1.0 m/s2. Do the trains collide? Hint for a solution: Determine how far each train would need to travel to come to a complete stop. Is the total distance less than 950 m? Unit conversions:

  21. Given information for the first train: vo = 20 m/s a = - 1.0 m/s2 v = 0 = rest or complete stop Dx = ? Given information for the second train: vo = 40 m/s a = - 1.0 m/s2 v = 0 = rest or complete stop Dx = ?

  22. If it takes 200 m for the first train to stop, and 800 m for the second train to stop, the combined necessary stopping distance is 1000 m. This is just a little more than the 950 m given for the trains to stop. OOPS!!!!

  23. Example 5: A car sits at rest at a red light. The moment the light turns green, a truck passes the car with a constant speed of 10.0 m/s. At the same moment, the car begins to accelerate at 2.50 m/s2. Assuming the car continues with a constant acceleration, how long will it take for the car to catch up to the truck? How far will they travel? How fast will the car be traveling when it passes the truck? Hint for a solution: The motion of the car and truck begin at the same place and time. The car and truck also are at the same place at the same time at the end of the motion. In other words, the car and the truck travel the same distance in the same time. Given information: vo truck = 10.0 m/s vo car = 0 acar = 2.50 m/s2 atruck = 0 Unknown information: Dx = ? t = ? vcar = ? Use the above equation twice, once for the car and once for the truck.

  24. First, for the truck… Second, for the car… Since these are the same Dx and t for both equations, set the two equations equal to one another. Solve for t ……

  25. One solution is t = 0. This is the beginning of the situation. Find the answer other than zero…. Divide both sides by t. Solve… Now substitute this time into one equation for Dx….

  26. Finally, the speed of the car can be found with: Note that the speed of the car is double the speed of the truck. This will always be true if the car starts from rest.

  27. Example 6: A rocket car accelerates from rest at a rate of 124 m/s2. (!!!) (a) How fast will the car be traveling at a time of 5.00 seconds? (b) How far will the car travel during its 5th second of motion? Given information: vo = 0 a = 125 m/s2 t = 5.00 s Unknown information: v = ?

  28. (b) How far will the car travel during its 5th second of motion? Given information: vo = 0 a = 125 m/s2 t = ? Unknown information: Dx = ? The 1st second lasts from t = 0 to t = 1 s. The 2nd second lasts from t = 1 s to t 2 s. Thus the 5th second spans from t = 4 s to t = 5 s. Subtract the displacement for 4 s from the displacement for 5 s.

  29. Example 7: A ping pong ball is shot at a rate of +20.0 m/s through a tunnel that is capable of producing a constant acceleration of –2.00 m/s2 on the ball. (a) At what time(s) will the ball be at the 50.0 m mark? Given information: vo = +20.0 m/s a = -2.00 m/s2Dx = +50.0 m Unknown information: t = ?

  30. The first time corresponds to the ball passing the 50.0 m mark as the ball is moving towards the right. Eventually the ball will slow to a stop. Since the wind is still pushing on the ball, it will start to accelerate towards the left, back towards its initial position. The second time corresponds to the ball passing the 50.0 m mark on the return trip towards the left.

  31. (b) At what time(s) will the ball be at the 200.0 m mark? Given information: vo = +20.0 m/s a = -2.00 m/s2Dx = +200.0 m Unknown information: t = ?

  32. This time does not exist, as the value under the radicand is negative. A real answer does not exist, so the ball must never reach the 200 m mark.

  33. (c) What is the velocity of the ball when it returns to the origin? Given information: vo = +20.0 m/s a = -2.00 m/s2Dx = 0 m Unknown information: v = ? Since the ball is moving towards the left on the return trip, take the negative sign to show this direction of motion. Thus:

  34. Example 8: Mr. Morris is riding his bike at 25 km/hr towards school when a squirrel leaps from behind a bus and stops 4.55 meters directly in front of him. Mr. Morris hits the brakes and decelerates at a rate of 2.80 m/s2. (a) Does he hit the squirrel? (b) If no, by what distance does he miss. If yes, how fast is he traveling when he hits? Hint: Solve for the distance to stop: Given information: vo = +6.944 m/s a = -2.80 m/s2 v = 0 m/s Unknown information: Dx = ?

  35. This is more than the 4.55 meters distance to the suicidal squirrel.

  36. (b) If no, by what distance does he miss. If yes, how fast is he traveling when he hits? Given information: vo = +6.944 m/s a = -2.80 m/s2Dx = 4.55 m Unknown information: v = ?

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