Linear Motion with constant acceleration Physics 6 A

As long as its acceleration is constant*, the motion of any object can be

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s

Sample Problem #2 A baseball is hurled straight up with an initial speed of

Sample Problem #3 A model rocket is fired vertically from a launch pad on

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Linear Motion with constant acceleration Physics 6 A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

As long as its acceleration is constant*, the motion of any object can be described by the following simple formulas: (1) (2) (3) When using these formulas it is important to be clear about your variables. Especially important is where and when the position and time are zero. A couple of examples will help clarify this. *both the magnitude and direction must be constant Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking By definition, average acceleration is just (change in velocity)/(change in time). Since the car sped up from 0 m/s to 30 m/s in 5. 6 seconds, this is a simple calculation: This is the answer for part a) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Another way to do part a) is to use formula (2) from before: In this case we are starting from rest, so v 0 = 0 m/s, and our final speed is v = 30 m/s. The elapsed time is t = 5. 6 s Putting these numbers into the equation we get: 30 m/s = 0 + a(5. 6 s) Solving for a gives us the answer: a ≈ 5. 4 m/s 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now that we have the acceleration from part a), we can get the answer to b) Which of our 3 equations is appropriate here? (1) (2) (3) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now that we have the acceleration from part a), we can get the answer to b) Which of our 3 equations is appropriate here? Equation (1) or (3) will work in this case. (1) (2) (3) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Using Equation (3): Note: The discrepancy is due to round-off error from the original calculation of a Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now we move on to the braking part of the problem: We can organize our information in a table if we want: x 0 = x= v 0 = v= Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now we move on to the braking part of the problem: We can organize our information in a table if we want: x 0 =0 x =50 m v 0 =30 m/s v= 0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now we move on to the braking part of the problem: We can organize our information in a table if we want: x 0 =0 x =50 m v 0 =30 m/s v= 0 Compare this information with our formulas from earlier. Which formula is the easiest one to use to find acceleration? (1) (2) (3) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now we move on to the braking part of the problem: We can organize our information in a table if we want: x 0 =0 x =50 m v 0 =30 m/s v= 0 Compare this information with our formulas from earlier. Formula (3) gives us what we need. (1) (2) (3) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now that we have acceleration, we can find the time for part d) Which formula should we use? (1) (2) (3) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #1 A 2009 Audi A 4 accelerates from rest to 30 m/s in about 5. 6 seconds. From that speed, it stops in a distance of 50 meters. Find the following: a) the average acceleration while it is speeding up b) the distance traveled while speeding up c) the average acceleration while braking d) the time elapsed while braking Now that we have acceleration, we can find the time for part d) Which formula should we use? Either of (1) or (2) will work, so choose the easier one: Formula (2) (1) (2) (3) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? Now we can think about the problem: What is special about the maximum height? In other words, do we know anything specific about the ball when it reaches the maximum height? YES! This is the most important piece of information in this type of problem: V 0 = 30 m/s y=2. 1 m the vertical velocity is zero at the top. y=0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? v = 0 here ymax=48 m Where is the ball when it is traveling at its maximum speed? Just before it hits the ground. A good way to think about this is to realize that it is speeding up as it falls, so the farther it falls, the faster it is moving. * V 0 = 30 m/s y=2. 1 m Fastest here Note: as the ball passes by the starting point it has a velocity of 30 m/s downward. This is just the opposite of the initial velocity! In fact, at any height, the ball will be moving at the same speed on the way up or the way down. *We are ignoring air resistance in this problem, so there is no terminal speed. y=0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? We need to put in y, y 0, v 0 and g We know it lands on the ground, so y = 0 If we use y 0 = 2. 1 m and v 0 = 30 m/s: V 0 = 30 m/s y=2. 1 m If we want to solve this one, we need to use the quadratic equation. This will work just fine, but it is more work than we should really need to do here. We get 2 answers: y=0 t = 6. 2 s and t = -0. 07 s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? We need to put in y, y 0, v 0 and g We know it lands on the ground, so y = 0 If we use y 0 = 2. 1 m and v 0 = 30 m/s: V 0 = 30 m/s y=2. 1 m If we want to solve this one, we need to use the quadratic equation. This will work just fine, but it is more work than we should really need to do here. We get 2 answers: y=0 t = 6. 2 s and t = -0. 07 s Prepared by Vince Zaccone Choose the positive answer in this case. For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? v = 0 here ymax=48 m That took some tedious algebra to find the answer. How can we avoid this? We can try using a different starting point. How about starting at the top? Putting in y 0 = 48 m and v 0 = 0: V 0 = 30 m/s y=2. 1 m This time it is much easier to solve! y=0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? v = 0 here ymax=? We get 2 answers again: t = 3. 1 s and t = -3. 1 s Again we choose the positive value, but does it actually answer our question? V 0 = 30 m/s y=2. 1 m y=0 NO! This is only the time it takes to fall from the maximum height. We need to add it to the time it took to get to that max height. We found that earlier – it was 3. 1 s So our answer is the total time of 6. 2 s, as before. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #2 A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2. 1 meters high. Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions: a) What is the maximum height reached by the ball? b) When does it achieve that maximum height? c) What is the maximum speed achieved by the ball? d) When does the ball hit the ground? v = 0 here ymax=? Now we can do it one more time, but using Equation (2) instead. We already found the final speed to be v = 48 m/s Here is the calculation: V 0 = 30 m/s y=2. 1 m IMPORTANT NOTE: make sure you put in the correct final velocity here. It is negative because the ball is moving downward! y=0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #3 A model rocket is fired vertically from a launch pad on the ground. The engine provides an upward acceleration of 5 m/s 2 for a time of 10 seconds. After that time the rocket is in free fall. Assume that the parachute fails to deploy, and the rocket eventually plummets to the ground. Sketch graphs of the rocket’s vertical height vs. time and its velocity vs. time. Find the maximum height of the rocket. When does it land, and how fast is it moving just before it hits the ground? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Sample Problem #3 A model rocket is fired vertically from a launch pad on the ground. The engine provides an upward acceleration of 5 m/s 2 for a time of 10 seconds. After that time the rocket is in free fall. Assume that the parachute fails to deploy, and the rocket eventually plummets to the ground. Sketch graphs of the rocket’s vertical height vs. time and its velocity vs. time. Find the maximum height of the rocket. When does it land, and how fast is it moving just before it hits the ground? Answers: Max Height = 377. 6 m Time to impact = 13. 9 s Impact speed = 86 m/s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB