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Right now you are traveling approximately:. A) 0 mph. Relative to guy next to you. B) 900 mph. Relative to earth’s axis of rotation. C) 67,000 mph. Relative to the sun’s center. D) 1.8 times the speed of light (335,000 mi/s) .

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## Right now you are traveling approximately:

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**Right now you are traveling approximately:**A) 0 mph Relative to guy next to you B) 900 mph Relative to earth’s axis of rotation C) 67,000 mph Relative to the sun’s center D) 1.8 times the speed of light (335,000 mi/s) Relative to the point on the opposite side of the expanding universe according to Hubble and Einstein E) All of the Above**All Motion is Relative!**• When is anything ever completely still? • How do you describe the motion of a turning wheel or a falling piece of paper? • How fast are you traveling if you are sitting in a supersonic airplane? • Unless otherwise specified, all motion is relative to the surface of the earth!**Distance v. Displacement**• Simply put, distance is the magnitude (size) of displacement. • 5.0 m away is an infinite number of locations (because of infinite directions) • 5.0 m, North is a single value! • Distance is a scalar quantity! • Displacement is a vector quantity!**Vector Quantity**Fully described by both magnitude (number plus units) AND direction Represented by arrows -velocity -acceleration -force Scalar Quantity Fully described by magnitude (number plus units) alone -mass -temperature Vector v. Scalar Quantities**Vector values are represented by arrows!**Adding vectors that lie along a straight line: 3 m, North + 4 m, North = 7 m, North 3 m, North + 4 m, South = 1 m, South Resultant vectors Head to tail method **Vector diagrams are not used when adding vectors in one**dimension (along a line): 3 m, North + 4 m, North = 7 m, North 3 m + 4 m = 7 m or -3 m + -4 m = -7 m 3 m, North + 4 m, South = 1 m, South 3 m + (-4 m) = -1 m or -3 m + 4 m = 1 m For a vector-- Sign does not represent value, it represents direction! Traditionally: Up/Right (+) Down/Left (-)**Vectors exist in one, two and three dimensions!**For now, we will only deal with vectors in one dimension…. …therefore the direction of the vectors will be represented by + and - signs Following the conventions of the coordinate axis system: Up/Right (+) Down/Left (-)**Speed and Velocity in one dimension**• Speed is the magnitude of velocity- it only reflects how fast an object is traveling • Velocity is a vector- it is speed in a particular direction • Along a straight line direction is indicated with a +/- sign • Again: Up/Right is positive and… • Down/Left is negative**Speed v. Velocity**A old man leaves his house, walks 1mile North in 1 hour, turns and walks 1 mile east in 1 hour, turns and walks 1 mile south in 1 hour, and then turns and walks 1 mile west in 1 hour. What was his average speed? What was his average velocity? Average speed is total distance divided by total time: v = ∆d/∆t = 4 mi / 4 hr = 1 mi/h Average velocity is total displacement divided by total time: v = ∆d/∆t 0 mi/h**Calculating Velocity**• Average velocity is the total displacement divided by the total time: vav = ∆d ∆t d – d0 t – t0 = ∆(delta) means “change in” t final (the last during that time period)t0 initial (the first during that time period) • Instantaneous velocity is the velocity of the object at any given instantGBS Physics. • Constant (uniform) velocity means non-changing (non-zero) velocity.**Equation Notations in Physics:**Quantities are expressed in variable form with a subscript used to describe different points in time v = d – d0 or d2 - d1 t – t0 t2 - t1 The bar over the variable means the average quantity– in this case speed/velocity Speed/Velocity are derived quantities and have derived units of length/time Standard unit: m/s sometimes km/h**1) You drive 18.0 km to school in 30.0 minutes. Because of**bad weather, it takes you 45.0 minutes to get home from school. What was your average velocity (km/h)? What was your average speed? 2) Two trains travel toward each other on parallel tracks at the same speed. At one point, they are 9.0 km apart and they meet 6.0 minutes later. What is the velocity of each train? 3) A remote control car zips 60.0 m in 6.0 s away from the remote driver, abruptly reverses and stops 4.8 s after reversing at a point 18 m in front of the driver. What was the average speed of the car? What was it’s average velocity?**1 .5 1**2 1.5 1 t0 3 2.5 1 t1 4 3.5 1 t2 Speed gain: 1 m/s per every second t3 t4**The Nature of Acceleration**• Acceleration is the change in velocity divided by the change in time- the rate at which the velocity changes-GBS Phys • If an object speed up, it accelerates! • If an object slows down, it accelerates! • If an object does not change speed, but changes direction, it accelerates! -this is the case when an object travels in a circular path at constant speed**Calculating Acceleration**• We are only going to be concerned with acceleration in a straight line-- FOR NOW! • Acceleration is a vector, so in a straight line the direction will be represented by +/- sign. • Negative acceleration means opposite direction! • Average acceleration is change in velocity divided by change in time. • Uniform acceleration mean constant, non-zero acceleration!**v - vo**∆t This is the basic definition of acceleration in the form of an equation. aav = m/s2unit of acceleration (meter per second per second) The following equations are basic equations which are derivations of our basic definitions of velocity and acceleration: v = vo + a∆t 1 2 ∆d = vo∆t + a∆t2 v = √vo2 + 2a∆d**A car accelerates from rest at a rate of 2.00 m/s2 for 8.50**s. How fast is the car moving now? How far did it move during this time? vo = 0 v = vo + a∆t a = 2.00 m/s2 = 0 + (2.00 m/s2)(8.50 s) ∆t = 8.50 s = 17.0 m/s v = ? ∆d = vo∆t + .5a∆t2 ∆d = ? = 72.3 m = .5(2.00 m/s2)(8.50s)2**1) An object is accelerated to 35.0 m/s at a rate of 2.00**m/s2 for a time of 7.34 s. How far did it move during this time? [203 m] 2) How far does an object travel if it accelerates from 12.0 m/s to 45.0 m/s in 15.3 s?[436 m] 3) An object traveling 48.7 m/s is brought to a stop in 7.61 s. How far did the object move during this time? [185m] 4) An object rolled up a hill will stop after having traveled up the hill 18.0 m. If the object is slowed at a rate of 2.50 m/s2, how long will it take to stop rolling? [3.80s]**1) A ball starting from rest rolls down an incline and**accelerates at the rate of 1.50 m/s2. The total length of the incline is 80.0 m. Will the ball reach the bottom after 10.0 s? 2) A car traveling at 35.0 m/s slows down at the rate of 1.50 m/s2 for 7.00 s. How far does the car travel in this time? 3) A bicyclist slows his bike from 25.0 m/s to 5.00 m/s over a distance of 75.0 m How long must this have taken?**Freefall**• An object whose motion is only affected by the acceleration rate due to gravity. • To keep it simple, we will ignore all the affects of air resistance on all objects in freefall. • With no air resistance, a potato chip will fall at the same rate as a bowling ball. • All objects in freefall will have the same acceleration-- the acceleration due to gravity: g = 9.80 m/s2**Freefall or Not Freefall?**An airplane in flight? Not Freefall-air and motors involved! A floating balloon? Not Freefall-floats on air A dropped rock? Freefall A ball thrown straight up? Freefall-up or down is irrelevant!**Solving Freefall Problems**• The acceleration is an accepted value of 9.80 m/s2 (does not count for SD!) • This acceleration will probably NOT be explicitly stated in the problem. • Traditionally, in Physics, up is positive and down is negative. • Therefore, for us, acceleration due to gravity will be a negative value whether the object is rising or falling!**A ball is launched vertically upward with a velocity of 115**m/s. A) To what height will it rise?B)How long will it take the ball to fall back to the earth? vf2 - vi2 2a A) ∆d = vo = 115m/s a= g = - 9.80m/s2 = 0 - (115m/s)2 2(- 9.80 m/s2) v = 0 A) ∆d = ? B) ∆ttotal= ? = 675 m (Positive meaning in the up direction!)**v - vo**a B) ∆t = v = - 115 m/s = -115 m/s - (115 m/s) - 9.80 m/s2 = 23.4 s Another method: v = 0 v - vo a B) ∆t = = 11.7 s X 2 = 23.4 s**1) A baseball is thrown down from the roof of a building and**it hits the ground 1.17 s later with a speed of 41.5 m/s. How far is the roof above the ground? 2) A batter pops a ball straight up with a speed of 30.0 m/s. How long does the catcher have to catch it, assuming he catches it at the same level as the bat? How high will the ball rise? 3) A student wants to throw a ball upward so that it reaches an exact height of 20.0 m above the point where he lets the ball go. With what speed must he throw the ball? How long will the ball be in the air?**4) A man holds a ball of mass 1.5 kg out over a sheer cliff**that is 75.0 m above the ground. The man throws the ball straight up with a speed of 25.0 m/s, but he fails to catch the ball as it falls back. With what velocity will the ball hit the ground? How long will the ball be in the air? 5) A hot air balloon is ascending with a speed of 4.00 m/s and is 102 m above the ground when it releases a sandbag. How much time do the people below the balloon have to get out of the way of that sandbag? End Notes**GBS Acceleration Demo**Which car(s) travel at constant speed? Which car(s) travel with acceleration? Which car(s) have the greatest acceleration? Back

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