2.4 Kinematic Equations

2.4 Kinematic Equations On pp. 46-47 of the text, four kinematic equations are derived: These all carry the assumptions that (1) acceleration is constant, and (2) t0 = 0. If the latter assumption is not the case, you can still use these equations, making the substitution t → t – t0. You will use these equations on countless occasions throughout PC141. Make sure to include them on your exam cheat sheets!

2.4 Kinematic Equations We have seen that the slope of a velocity – vs – time graph is equal to the instantaneous acceleration. We can gain more information by examining the area under the line. In the top graph on the right, , and . The area is that of a triangle with base t and height v. This area is . But (from the 3rd equation on the last slide) this is equal to . The same result is obtained even when (bottom plot), as you should check for yourself.

2.4 Kinematic Equations As it turns out, the area under the v(t) curve is equal to even when acceleration is not constant. However, to prove this (or to apply it in problem solving) requires calculus, so we won’t worry about it. The same approach can be used to determine using an acceleration – vs – time graph. Later in today’s class, we will tackle a problem that involves piecewise continuous acceleration.

Problem #1: Average Speed WBL LP 2.15

Problem #2: From the Earth to the Moon In his 1865 novel From the Earth to the Moon, Jules Verne proposed sending people to the moon via a rocket cannon that accelerates a capsule along a length of 220 m, attaining a final speed of 11.2 km/s. What acceleration would the passengers experience? Fun(ish) fact: 11.2 km/s is the “escape velocity” from Earth. An object must attain this velocity if it is to break free from the gravitational pull of Earth. Solution: In class

Problem #3: A Blow to the Head A soccer ball is headed with and without the use of a helmet. In both cases, the acceleration of the player’s head is plotted as a function of time. At time t = 7 ms, what is the difference in the speed acquired by the bare head and the speed acquired by the helmeted head? • Solution: In class

Problem #4: Sprinting The current world record for the 100 m dash is 9.58 seconds, set by Usain Bolt in 2009. Analysis of that run shows that he reached his top speed around the 70 metre mark, at which point his speed was 12.42 m/s. What was his average speed during the race? Assuming that his acceleration was constant over the first 70 metres, what was the magnitude of his acceleration during that portion of the race? Solution: In class

Problem #5: Worst. Car. Ever. The “slowest” car ever produced is the 2009 Tata Nano, which accelerates from 0 to 60 miles per hour in a blistering 29.4 seconds. What is the magnitude of this acceleration, assuming that it is constant over this duration? How much distance does it cover during this time? Solution: In class

Problem #6: Sprinter vs. Car It seems that the 2009 Tata Nano has a much weaker acceleration that Usain Bolt. How much of a head start would the car require in order to “beat” Usain Bolt over 100 metres? Solution: In class

2.5 Free Fall The most obvious example of constant acceleration occurs when an object moves near the Earth’s surface (by “near” we mean anywhere within ±100 km or so of sea level). Here, there is a constant acceleration directed downward, with a magnitude of The symbol “g” indicates that gravity is responsible for this acceleration…but you didn’t hear that from me, since we’re not discussing dynamics (the why of motion) until chapter 4. In chapter 7 we will learn where the number “9.80” comes from. For now, just know that it is affected not only by altitude, but also by latitude and longitude, for various reasons. Therefore, it is not useful to quote g with any more significant figures than that which is shown above.

2.5 Free Fall When an object’s motion is influenced only by g, we say that it is in free fall. Although this term seems to imply that the object is moving downward, it in fact refers to any motion that is subject only to g, even if it projected upward or at an angle (such as a drive of a golf ball). Remember, even when an object is traveling upward, it is still accelerating downward. This is what causes it to slow down, momentarily stop, then begin its downward journey. Notice that g doesn’t depend on any properties (mass, shape…) of the moving object. This is only true in a vacuum. Otherwise, air resistance comes into play, and the object isn’t truly in free fall.

2.5 Free Fall When analyzing objects as they move vertically, it is customary to label the vertical axis as y, and to let y increase in the upward direction. In this case: • We can still use all of the 1D kinematics equations from this chapter, replacing x with y. • When applying these equations, use an acceleration . The negative sign indicates that acceleration is in the negative y direction. If you forget the negative sign, you’re essentially claiming that your object is “falling up”!

2.5 Free Fall Here are a few facts about free fall motion for an object thrown vertically upward. These may be of benefit in solving more difficult problems later on… • If the object returns to its original elevation, the upward and downward portions of the journey take the same time. • If the object returns to its original elevation, it does so with the same speed as that at which it was launched. The velocity isn’t the same, since the direction has changed. This concept will be beneficial when we study kinetic and potential energy. • At the top of the object’s trajectory, its velocity is momentarily zero.

Problem #7: Dropping Rocks

Problem #8: Your Idiot Upstairs Neighbour WBL EX 2.73 A particular apartment building has floors which are vertically separated by 4.0 m. Somebody from an upper floor drops an apple from their window. A resident on the 1st floor measures the time that the apple takes to pass across his entire window, which measures 1.35 m vertically. From what floor was the apple dropped? Solution: In class

Problem #9: Vertical Rocket WBL EX 2.75 A model rocket is fired vertically upward, with a constant acceleration of 12.0 m/s2. When it reaches a height of 1000 m, the engines immediately cut off, leaving the model rocket in free fall. What is the rocket’s speed at the moment the engines cut off? What is the maximum altitude of the rocket? What is the time it takes to reach this maximum altitude, measured from the moment of launch from the ground? Solution: In class

2.4 Kinematic Equations