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A skydiver is falling down (y component negative) at a speed of 48 m/s. She opens her chute. 11 seconds later, she has slowed down to 26 m/s. Assume her acceleration is constant. Draw graphs for: a. Acceleration (numerically correct) b. Velocity (numerically correct) c. Position (just the shape – assume the ground is at zero) What will the fan cart do when started with these parameters (a = +2m/s2, v0 = -5m/s)? A. Speed up to the left B. Speed up to the right C. Slow down to the left D. Slow down to the right Draw the graphs for your prediction (x0 = 0) One dimensional constant acceleration http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04 70223553&assetId=160341&resourceId=15300 Motion with Constant Acceleration A jet plane has an acceleration of 2m/s/s, starting from rest, at the origin. Fill in the table. Don’t use kinematic equations: t (s) 0 1 2 3 4 velocity (m/s) Motion with Constant Acceleration A jet plane has an acceleration of 2m/s/s, starting from rest, at the origin. After 1 s, what is the position of the object? t (s) velocity, v (m/s) position, x (m) 0 0 0 1 2 X A. 2m B. 1 m C. Somewhere between 1m and 2m D. Can’t be determined without more information Motion with Constant Acceleration (v0 v) t x 2 Δt is the time elapsed between v0 and v1. Motion with Constant Acceleration A jet plane has an acceleration of 2m/s/s, starting from rest, at the origin. Fill in the table. t (s) velocity, v (m/s) 0 0 1 2 2 4 3 6 4 8 position, x (m) Constant Velocity vs. Constant Acceleration http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04 70223553&assetId=160341&resourceId=15300 After 1 second, which ball has gone further, the one with constant velocity (red) or the one with constant acceleration? Constant Velocity vs. Constant Acceleration http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04 70223553&assetId=160341&resourceId=15300 Go to www. Wiley.com, physics and astronomy, introductory physics(algebra-based) keyword cutnell Go to student companion site Mute the audio please!! Prove the following, using math: At what time do the balls meet? At what position? Pumping Blood – Example #1 The left ventricle of the heart, each time it pumps, accelerates blood from rest to a velocity of 26 cm/s. It takes 0.15 s for the heart to go through one cycle. How far (2 sig figs) does the blood travel? a. 3.9 cm b. 2.0 cm Pumping Blood - Example #1 The left ventricle of the heart, each time it pumps, accelerates blood from rest to a velocity of 26 cm/s. It takes 0.15 s for the heart to go through one cycle. How far (2 sig figs) does the blood travel? 2.0 cm What is the acceleration that the heart provides? Example Problem #2 – turning point Bob throws the ball straight up at 20 m/s releasing the ball 1.5 m above the ground. a. What is the maximum height of the ball above the ground? b. What is the ball’s impact speed when it hits the ground? Draw velocity, acceleration and position graphs Example Problem #2 – turning point Bob throws the ball straight up at 20 m/s releasing the ball 1.5 m above the ground. a. What is the maximum height of the ball above the ground? b. What is the ball’s impact speed when it hits the ground? Step 1: pictorial representation, using particle model. Pictorial Representation – Draw an x or y axis and label the origin. Right/up are positive directions. This is a position, not time axis. – Put a dot at each important point in the problem, i.e. the beginning, the end, each change of acceleration, turning point (v=0). – List the kinematic variables (position, time, velocity) at each dot. At the first dot, x0, v0, t0, at the second dot, x1, v1, t1, and so on. – Show a0 between the 0 variables and the 1 variables with an arrow showing the appropriate direction of acceleration. Repeat as required. Bob throws the ball straight up at 20 m/s releasing the ball 1.5 m above the ground. Y axis with the ground chosen to be the origin (arbitrary decision) Three important points: ball release (y0), high point (y1), impact point (y2), which happens to coincide with the origin. Acceleration due to gravity on the way up and down so no change in acceleration. It’s not necessary to draw Bob! 2. List “knowns” and quantities you need to find. Knowns Find y0 = 1.5m y1, v2 v0 = 20 m/s t0 = 0s (beginning of story) a0 = -9.8m/s2 v1 = 0 m/s (high point) y2 = 0 m (as we defined it!) a. What is the maximum height of the ball above the ground? b. What is the ball’s impact speed when it hits the ground? 3. Decide which equation to use to solve part a. Does the object have constant v? No. Is time important (a “known” or a find)? No. Use the equation without time: v2 = v02 + 2a (∆y) Knowns y0 = 1.5m v0 = 20 m/s t0 = 0s a0 = -9.8m/s2 v1 = 0 m/s y2 = 0 m Find y1, v2 Bob throws the ball straight up at 20.0 m/s releasing the ball 1.5 m above the ground. a. What is the maximum height of the ball above the ground? For part a, is time important? The answer is no, so use equation #3 with appropriate symbols: v12 = v02 + 2a0∆y1,0 where ∆y1,0 = y1- y0 Solve for ∆y1,0 = - v02/2a (negative from moving v02 to the left side of equation) ∆y1,0 = 20.4 m, so y1 = 21.9 m. Knowns y0 = 1.5m v0 = 20 m/s t0 = 0s a0 = -9.8m/s2 v1 = 0 m/s y2 = 0 m Find y1, v2 Bob throws the ball straight up at 20.0 m/s releasing the ball 1.5 m above the ground. a. What is the maximum height of the ball above the ground? b. What is the ball’s impact speed when it hits the ground? For part b, is time important? The answer is no, so use equation #3 with appropriate symbols: v2 = v1 + 2a0∆y2,1 where: 2 2 ∆y2.1 = y2 - y1 = -y1 Solve for v2 = +/- (- 2a0y1)1/2 v2 = -20.7 m/s (it’s negative b/c ball is going down Assess: reasonable , since it fell just a tad more than when it was launched. Knowns y0 = 1.5m v0 = 20 m/s t0 = 0s a0 = -9.8m/s2 v1 = 0 m/s y2 = 0 m y1 = 21.9 m Find v2 How high does it go? A coin is tossed into the air with an initial velocity of 8.0 m/s. How high does it go before starting to fall? Does it take the same amount of time to fall back to it’s starting point as it took to go up? How can you show this? We usually use “y” for vertical motion How high does it go? http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04 70223553&assetId=160341&resourceId=15300 When different accelerations are involved, you must divide the problem into parts – one for each different acceleration A “rocket sled” starts at rest and accelerates at 50 m/s/s for 5.0 s. It runs out of fuel and coasts for 3.0 s, with no change in speed. A braking parachute deploys and the sled slows to a stop, accelerating at a rate of 3 m/s/s to do so. A. What is the maximum velocity of the sled? B. What is the total displacement of the sled? Draw a pictorial representation for this problem with the correct number of important points and list knowns and “finds” with appropriate subscripts. Acceleration should have correct signs (use diagram if not sure). A “rocket sled” starts at rest and accelerates at 50 m/s/s for 5.0 s. It runs out of fuel and coasts for 3.0 s, with no change in speed. A braking parachute deploys and the sled slows to a stop, accelerating at a rate of 3 m/s/s to do so. Find the max velocity and the total displacement of the sled. Note difference between t2 (8s after t0) and ∆t2 (t2 - t1), which is 3s. v 1 = v2 0 Rocket sled problem A. What is the maximum velocity of the sled? 0 Time is important. Maximum velocity occurs in the first segment, where sled speeds up: vmax = v1 v1 = v0 + a0(t1-t0) = 250 m/s v 1 = v2 Rocket Sled Problem B. What is the total displacement of the sled? Time is important for the first 2 segments. Find final position in each segment with different acceleration: x1 = x0 + v0 (t1-t0) + ½ a0 (t1-t0)2 x1 = 625 m x2 = x1 + v1 (t2-t1): constant v x2= 1375m No time is given for the 3rd segment: v32 = v22 + 2a2(x3 - x2) x3 = 11,800 m, which is the total displacement. v 1 = v2 Rocket sled problem, answers • vmax = v1 = 250 m/s • Δxtot = (x3 – x0) = x3 = 11,800 m Different accelerations A car starts from rest at a stop sign. It accelerates at 4.0 m/s/s for 6.0s, coasts for 2.0 s, and then slows down at a rate of 3.0 m/s/s to come to a complete stop right at the next stop sign. How far apart are the stop signs? • Draw a pictorial representation, choose a zero, list knowns and “finds” and determine whether time is important for each segment of the motion. Different accelerations - answer A car starts from rest at a stop sign. It accelerates at 4.0 m/s/s for 6.0s, coasts for 2.0 s, and then slows down at a rate of 3.0 m/s/s to come to a complete stop right at the next stop sign. How far apart are the stop signs? Answer: Δxtot = x3 = 216 m. Time was important in the first segment, and we need to use both position and velocity equations to get v1 and x1. The next segment involved constant velocity, so there is only one equation to choose. The 3rd segment did not involve time so we could choose the “timeless” equation. Free Fall Problem (#52) • A pellet gun is fired straight downward from the edge of a cliff that is 15 m above the ground. The pellet strikes the ground with a speed of 27 m/s. How far above the cliff would the pellet have gone if the gun had been fired straight upward? (Note: the bullet has the same initial speed whether fired up or down). • For 2 object problems, there is always at least one quantity that is the same for both objects. What is it for gun shooting up and gun shooting down? There may be more than one! Use your pictorial representation and list of variables to figure out the important quantity or quantities. HardFree Fall Problem (#52) v0D = -v0U = -20.9 m/s. How far the bullet goes is y1U = 22.3m 2.6 Freely Falling Bodies • If the only acceleration acting on an object is the acceleration due to gravity, the object is said to be in freefall . • Acceleration due to gravity is a vector with a magnitude of 9.8 m/s/s and a direction towards the center of the earth (“true vertical”). An object in freefall will change its velocity by just under 10 meters per second every second if it is traveling towards or away from the center of the earth. For this chapter, that means up or down. • The direction in which the gravity vector points is the operational definition of “true vertical”. • If on your paper, “down” is a negative direction (and it often is!) the only non-zero component of gravitational acceleration in negative. • Free fall refers to objects going up or down, with an initial velocity or not, if the only acceleration is due to gravity. Kinematic Equations for Constant Acceleration • x = x0 + vconst (∆t) (equation of a line for motion with constant velocity; this is the only equation of motion if a = 0). • v = v0 + a(∆t) (equation of a line for motion with constant acceleration; use if time is important). • x = x0 + v0 (∆t) + ½ a (∆t)2 (quadratic equation, use if time is important) • v2 = v02 + 2a (∆x) (Use this equation if time is not important). • ∆x = ½(v0 + v) (∆t) ( where ½(v0 + v) is vavg). All velocities are instantaneous unless noted Analysis of a position graph In which segment does the velocity have the greatest value? A. B. C. D. Not enough information to determine. Analysis of a position graph In which segment does the velocity have the greatest value? A. B. C. D. Not enough information to determine. Analysis of a position graph In which segment is the motion in a negative direction? A. B. C. D. Both B and C Making a velocity graph from a position graph Draw a numerically correct velocity-time graph for this positiontime graph. Motion graph for constant acceleration http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04 70223553&assetId=160341&resourceId=15300 Predict the graphs for position vs. time and velocity vs time for the spaceship.