Topics for the test and Sample Problems

Transcription

1 Topics for the test and Sample Problems 1. Be able to Rearrange Every Motion Equation on the Equation Page and Be able to use GUESS to choose correct equation. See the notes from 9/13 and 9/1 x = v t x f = x i + v t v=a t v f = v i + at x f = x i + v it + ½ at v f - v i = a (x f-x i) Δx = (v f+v i ) t x= ½(gt ) same as t = d g d x = v x t t = t = v yi g

2 Position (m) Height (m). Be able to describe motion based on PT Graph shapes. What are initial and final positions, is initial velocity positive, negative or zero? Is motion constant, accelerated or zero? 1 A) Accelerated motion. You can tell because the slope (velocity) is changing. The position starts at zero, but the slope does not so initial velocity is not zero. IN fact it is positive, and the final slope is negative. The slope at the peak is zero, so the velocity is zero there. Because velocity decreases throughout, the acceleration is negative throughout B) Constant motion. You can tell because the slope is constant. (left) Position starts at zero, but the slope does not. It is a constant negative at all points. Acceleration is zero. (right) ) Position starts at zero, as does slope. It is a constant positive at all points. Acceleration is zero C) Accelerated motion. You can tell because the slope (velocity) is changing. The position starts at zero, and so does the slope. It quickly becomes positive, and the final slope is even more positive. Because velocity increases throughout, the acceleration is positive throughout.. 3. Be able to describe motion based on PT Graph shapes There are three sections. In the first velocity is constant and positive (m/s). In the middle, the object is not moving. In the third segment the velocity is negative (1m/s). Acceleration happens at 3s and s. Both accelerations are negative.

3 Depth (ft). Be able to describe acceleration based on VT graph shapes. Pretend the left and bottom sides of the boxes are the y and x axes. A) Velocity is increasing, so acceleration is positive. Velocity starts at zero. (Goes with C) B) Velocity is decreasing, acceleration is negative. Velocity starts positive and decreases. (Goes with A) C) Velocity does not change, but remains at some constant positive number. Acceleration is zero. (Goes with B). Be able to determine a linear equation using the slope, y intercept and units Depth of pool based on distance from edge Be able to find slope and put it into a specific y=mx+b format Use correct labels and units First find the slope with units (1.ft/yd) Write the units on the y axis (Depth(ft)) Write the Units of the x axis (Distance from edge (yd)) Write the y intercept with units (.ft) Substitute these values into y=mx+b, keeping all the units! 1 3 Distance from edge (Yd Depth (ft) = Distance from edge(yd) *1.(ft/yd) +.ft

4 . Be able to find the position, time or velocity for a free falling object with any value of g On a planet with a gravitational acceleration of.3m/s, how long will it take a bowling ball to fall 3.m? Use GUESS G: a=.3 m/s d=3.m U: E: t = d g S: t = 3.m.3 m s = 1.1s What will its velocity be when it hits? Use GUESS G: a=.3 m/s d=3.m v i= t=1.1s U: v f E: v f=v i + at S: v f=m/s +.3m/s *1.1s =.1m/s (don t need d) 7. Be able to find the position, time or velocity for a free falling object with any value of g If Wylie Coyote runs off a cliff in Arizona at m/s, and travels meters before he realizes what he has done, how far below the level of the cliff will he be when he holds up his little sign? Use GUESS this is a tower problem G: V H U: d y (but you need to find time to find it) v iy m/s Vx = m/s v fy a=g= -9.8m/s NONE d y= m t= t= Finding time E: t=d/v S: d = m m = 1.s s Finding the free fall distance E: t = d g or d = ½at S: d = ½( 9.8 m s ) 1.s =-7.m 8. Be able to find hang time and maximum height for any object launched upward The rocket club has a new air rocket that will launch at.m/s. They launch it straight up. How high will it go and how long will it take to come down, if it does not have a parachute? Height. options you can find time first and then go back and calculate height, or calculate the values independently. Either way, use GUESS Without time G: a= -9.8 m/s v i=.m/s v f= U: distance (m) E: v f v i = adrearrange to v f v i = d S: Finding time first G: a= -9.8 m/s v i=.m/s v f= U: time(s) a m/s m s ( 9.8 m s ) = d =.m E: v f=v i + at rearrange to v f v i a = t S: Then find distance, using the time to go up m/s m s ( 9.8 m s ) = t =.s This is the time to go up, multiply by to get the total time G: a= -9.8 m/s v i=.m/s t=.s x i= U: distance (m) E: x f = x i + v it + ½ at S: x f = m s.s + ½ ( 9.8 m s ). s =.m

5 9. Be able to find hang time and maximum height for any object launched upward The rocket club has a new air rocket that will launch at.m/s. They launch it at an angle of 7. How high will it go and how long will it take to come down, if it does not have a parachute? This is just like the previous problem, except the initial vertical velocity must be calculated. v iy=v i sin θ = m/s* sim 7 =18.8m/s Height. options you can find time first and then go back and calculate height, or calculate the values independently. Either way, use GUESS Without time G: a= -9.8 m/s v i= 18.8.m/s v f= U: distance (m) E: v f v i = adrearrange to v f v i = d S: Finding time first G: a= -9.8 m/s v i= 18.8.m/s v f= U: time(s) a m/s 18.8 m s ( 9.8 m s ) = d = 18.m E: v f=v i + at rearrange to v f v i a = t S: Then find distance, using the time to go up m/s 18.8 m s ( 9.8 m s ) = t = 1.9s This is the time to go up, multiply by to get the total time G: a= -9.8 m/s v i= 18.8.m/s t= 1.9s x i= U: distance (m) E: x f = x i + v it + ½ at S: x f = 18.8 m s 1.9s + ½ ( 9.8 m s ) 1.9 s =18.m 1. Be able to find the value of g, based on a free falling object An astronaut drops a hammer on a planetoid. If the hammer falls 1.8m and takes.3s to complete the fall, what is the acceleration due to gravity on the planetoid? USE GUESS G: d=1.8m t=.3s v i= x i = U: a (g for the planet) E: x f = x i + v it + ½ at rearrange to a= x f/t S: a = 1.8m (.3s) =.8 m s

6 V (m/s) Height (m) 11. Be able to combine data from two graphs to solve a problem 3 1 PT for Up and Down What is the velocity when the object is on its downward trajectory and 1.m high? Read the PT graph for time be as accurate as possible (1.s). What is the velocity at this time? (-7m/s) What is the position when the velocity is zero? Read the VT graph for time or apply your knowledge about projectiles either gives a time of.9s, and the position is m. What is the position when the velocity is +.m/s? Read the VT graph for time be as accurate as possible (.s). What is the position at this time? (3.m) 1 VT for Up and Down Recognize Constant and Accelerated Motion from Data Sets. Determine if each is constant or accelerated, and if the motion is positive or negative Set 1 Set Set 3 Set t(s) d(m) t(s) d(m) t(s) d(m) t(s) d(m) Distance increases in positive direction every time. Increasing + velocity + acceleration Distance constant in negative direction every time. Constant negative velocity acceleration Distance increases in positive direction every time. Increasing + initial velocity, - final velocity + acceleration Distance constant in positive direction every time. Constant positive velocity acceleration

7 13. Be able to rank values from the greatest to least, or least to greatest. Rank the time needed to stop for these three cars v f=vi+at v f=rearrange to t= -v i/a 1) Initial velocity m/s, acceleration -m/s t= 1.s Greatest ) Initial velocity m/s, acceleration -m/s t=1s Least 3) Initial velocity 3m/s, acceleration -.m/s t=1s Middle 1. Be able to solve word problems. ALWAYS USE GUESS A cyclist pedals at a constant speed of.8m/s for 8s. How far do they travel? G: v=.8m/s t=8s U: distance (m) E: d=vt S:.8m/s*8s=78m A cat skitters to a stop. It takes 3.s to go from 8.m/s to m/s. What is the cat s acceleration? G: v i= 8.m/s v f= t= 3.s U: acceleration (m/s ) E: v f=v i + at rearrange to - v i/t= a S: -8.m/s 3.s= -.8m/s Compare the distances travelled by a car that gets to a light just as it changes to one that has to start from m/s The car that cruises to the light has a velocity of 7. m/s as it enters the intersection and accelerates at m/s. The car that has stopped accelerates at 3m/s. How far will each have travelled in 1..s? Car starting from m/s G: a= 3.m/s v i=.m/s t= 1.s x i= U: distance (m) E: x f = x i + v it + ½ at S: x f = m s 1.s + ½ (3 m s ) 1. s =1m Car starting from 7.m/s G: a=.m/s v i= 7.m/s t= 1.s x i= U: distance (m) E: x f = x i + v it + ½ at S: x f = 7. m s 1.s + ½ ( m s ) 1. s =17m