UAM Paradigm Lab. Uniform Acceleration Background. X-t graph. V-t graph now. What about displacement? *Displacement method 2 9/18/2017

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1 9/8/07 UAM Paradigm Lab Uniform Acceleration Background Wheel down a rail Observations Dots got further apart as the wheel rolled down rail This means the change in position increased over time X-t graph V-t graph now The slope of a position-time graph is the velocity. If the position-time graph is curved, the slope of a line tangent to the curve tells you the velocity at that time. The velocity at a given time is called the instantaneous velocity. You can now plot the instantaneous velocity at different moments in time (a clock reading) and plot that. acceleration is the rate of change in velocity, which is the slope of a velocity-time graph. Mathematically, Write the equation of the line v f = a*t + v o What about displacement? Two methods Method -make a new graph Plot position vs time, instead of time, analysis reveals the slope is now ½ of acceleration If x o is 0m, then: If x o is not 0m then: *Displacement method The area under a vt graph is displacement. For a triangle, like at right, the equation would become x = ½ v*t (if starting position is also 0m) We also know the equation of the line to be: v f= a*t + v o or v = a*t We can sub in the nd equation ( v = a*t) into v of the first equation and get x = ½ (a*t)t, now simplify it = ½ at BUT, what if there s a starting velocity? = ½ at + v o *t Equation looks super confusing, but this has the separated out into starting position and final position

2 9/8/07 Area = Displacement (a similar explanation) Important Formula! = vit + vt = vit + att = vit + at Which graph(s) do I use? I m a fan of using the v-t graph as much as possible. You can analyze 3 things on a graph (vtgraph) Slope (acceleration) Y-intercept (v o ) Area under graph (displacement) From the v-t graph you can get the only two equations you will need (95% of the time) v f = a*t + v o x f = ½ at + v o *t + x o (if v o and x o not zero) Most of the time you won t use an equation anyway, you will just look at points on the graph and do basic graphical analysis to solve a problem. The equations are there to help verify. What if I don t know time? Now you graph velocity vs position, another side-open parabola. All of these equations require time. (Below is a simplified explanation of the last slide) By combining two or more equations we can derive some that don t use time. This one is useful: Graph v vs x and now it s linear v f = (a)(x) +v Important Formula! v f i = v + a Velocity-Time Graphs

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4 9/8/07 The slope of a velocity time graph tells us how fast the speed is changing. This value is called acceleration. Find the acceleration of this car. Find the acceleration of this car. Calculating Acceleration Formula a = v t vf a = vi t rise/run = 5-5 / 5-0 = 0 / 5 = 4 m/s/s A car increases from m/s to 8 m/s in 5.0 sec. Calculate the acceleration. v 8 a = = = t m = s 4

5 9/8/07 Calculating Acceleration Consider this graph. Formula a = vf vi t A car changes from 8 m/s to m/s in 5.0 sec. Calculate the acceleration. v 8 60 m a = = = = t s How could we predict what the velocity at some time t would be? Important Formula Generally y = m x + b v f = vi + at Specifically vf = a t + vi A car starts from rest and accelerates at m/s/s for 0 seconds. How fast will it be going? Given: vi=0 a= m/s/s t=0 seconds A car starts at 80 m/s and slows down with an acceleration of -0 m/s/s for 4 seconds. What is its new velocity? Given: vi=80 m/s a= -0 m/s/s t=4 seconds Solution: vf = vi + at vf = 0 + ()(0) vf = 0 m/s Solution: vf = vi + at vf = 80 + (-0)(4) vf = 80 + (-40) = 40 m/s 5

6 9/8/07 The area under the curve of a velocity time graph tells us how far the object has traveled. This value is called distance if the signs are ignored and displacement if the signs are considered. Area for constant velocity graphs. This car is traveling at a constant speed. The area shaded is a rectangle. A = l x w = 6 x 30 = 80 m This car traveled 80 m. Area for changing velocity graphs. Area for changing velocity graphs. This car is traveling at a changing speed. The area shaded is a triangle. A = ½ x b x h = ½ x 4 x 40 = 80 m This car traveled 80 m. This shape is a rectangle plus a triangle. 0x3=60 ½ x 3x 30= =05 m Distance vs. Displacement Distance (average speed) add the absolute values =5m Displacement (average velocity) keep the signs. (+60)+(+45)+(-0)= (+85)m Freefall How fast was that pumpkin going? 6

7 9/8/07 We have been studying accelerations changing speeds. One of the most common accelerations is the one produced by gravity. We used gravity to accelerate our carts down an incline. They went faster and faster as they went along. Gravity is a force (a push or pull). Forces cause accelerations. Consider a book and a marble. If dropped together, which will hit the ground first? Why? 7

8 9/8/07 Which one does gravity pull harder on (which is heavier)? Which is more sluggish resisting changes in motion, harder to move? Let s do it! What if we use paper instead of the marble? Famous experiment feather and guinea. Apollo 5 8

9 9/8/07 Definition: Freefall and Falling Freely Acted on by gravity alone. No other force no rockets, no wind, no ropes, no parachute. Lesson: The acceleration due to gravity is the same for all object. g = 9.80 m/s a g = 9.80 m/s How fast? A rock falls from rest for 6.5 seconds. How fast will it be going? vf = vi + at vf = 0 + (9.80)(6.5) vf = 63.7m vf = 64m How fast? A rock is thrown downward with a speed of 5m/s and falls for 6.5 seconds. How fast will it be going? vf = vi + at vf = 5 + (9.80)(6.5) vf = m vf = 79m How fast? A rock is thrown upward with a speed of 5m/s and falls for 6.5 seconds. How fast will it be going? vf = vi + at vf = 5 + (9.80)(6.5) vf = m vf = 49m A rock falls from rest for 6.5 seconds. How far will it fall? How far? = vi( t) + at = 0 + (9.80)(6.5) = 07m = 0m 9

10 9/8/07 How far? A rock is thrown downward with a speed of 5m/s and falls for 6.5 seconds. How far will it fall? = vi( t) + at = (5)(6.5) + (9.80)(6.5) = = 30m How far? A rock is thrown upward with a speed of 5m/s and falls for 6.5 seconds. How far will it fall? = vi( t) + at = ( 5)(6.5) + (9.80)(6.5) = = 0m How far? A rock is thrown upward with a speed of 5m/s. How high will it go? vf = vi 0 = 5 + a + (9.80) 65 = = 3m (9.80) A ball is thrown upward with a speed of 0 m/s. Which takes longer, going up or coming back down? The motions symmetrical. The upward motion takes the same time as the downward motion. 0