9-11-96
Relevant sections in the book : 2.4, 2.5, 2.6, and 2.7
When the acceleration of an object is constant, calculations of the distance travelled by an object distance, the velocity it's travelling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined.
Note that the equations apply under these conditions: (1) the acceleration is constant (2) the object is assumed to start at x = 0 at t = 0 (3) the equations are vector equations, but the variables are not normally written in bold letters. The fact that they are vectors comes in, however, with positive and negative signs.
The equations are:
Note that the equations above are all derived in section 2.4
Doing a sample problem is probably the best way to see how you would use the equations. Let's say you're driving in your car, and approaching a red light on Commonwealth Avenue. A black Porsche is stopped at the light in the right lane, but there's no-one in the left lane, so you pull into the left lane. You're travelling at 40 km/hr, and when you're 15 meters from the stop line the light turns green. You sail through the green light at a constant speed of 40 km/hr, and pass the Porsche, which is accelerating at a constant rate of 3 m/s2.
(a) How far from the stop line do you pass the Porsche?
(b) When does the Porsche pass you?
(c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/hr.
Step 1 - Write down everything you know. This involves choosing a positive direction. In this case, let's go with positive being anywhere beyond the stop line, and negative being anywhere behind the stop line. Define an origin - the stop line is a good choice in this problem. Decide on a system of units...meters and seconds is a good choice here, so convert your speed to m/s from km/hr. Drawing a diagram is also a good idea.
Origin = stop line
Step 2 - Figure out what you need to solve for. At the instant you pass the Porsche, the x values (yours and the Porche's) have to be equal. You're the same distance from the stop line, in other words. Write out the expression for your x-value and the Porche's. We'll use the equation:
Because you don't start at the stop line when the light changes (and when the imaginary stopwatch keeping track of time starts going), we have to modify the equation a little to account for your starting position. The equation becomes:
For you : x = -15 + 11.11 t + 0
For the Porsche : x = 0 + 0 + 1/2 (3) t^2 = 1.5 t^2
At some time t, when you pass the Porsche, these x values will be the same. So, we can set the equations equal to one another and solve for time, and then plug the time back in to either x equation to get the distance from the stop line. Doing this gives:
-15 + 11.11 t = 1.5 t^2
Bringing everything to one side gives:
1.5 t^2 - 11.11 t + 15 = 0 This is a quadratic equation, which we can solve using the quadratic formula:
where a = 1.5, b = -11.11, and c = 15
This gives two values for t, t = 1.776 s and t = 5.631 s.
What do these two values mean? In many cases only one answer will be relevant, and you'll have to figure out which. In this case both are relevant. The smaller value is when you pass the Porsche, while the larger one is when the Porsche passes you back.
To get the answer to question (a), plug t = 1.776 into either of your x expressions. They should both give you the same value for x, so you can use one as a check.
For you, at t = 1.776, x = 4.73 m.
For the Porsche, at t = 1.776 s, x = 4.73 m.
We've actually already calculated the answer to (b), when the Porsche passes you, which is at t = 5.6 s.
To get the answer to part (c), we already know that you're travelling at a constant speed of 40 km/hr, which is under the speed limit. To figure out how fast the Porsche is going at t = 5.631 seconds, use:
v = vo + a t = 0 + (3) (5.631) = 16.893 m/s.
Converting this to km/hr gives a speed of 60.8 km/hr, so the driver of the Porsche is in danger of getting a speeding ticket.
Objects falling straight down under the influence of gravity are excellent examples of objects travelling at constant acceleration in one dimension. This also applies to anything you throw straight up in the air which, because of the constant acceleration downwards, will rise until the velocity drops to zero and then will fall back down again.
The acceleration experienced by a dropped or thrown object while it is in flight comes from the gravitational force exerted on the object by the Earth. If we're dealing with objects at the Earth's surface, which we usually are, we call this acceleration g, which has a value of 9.8 m/s2. This value is determined by three things: the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant. We'll be dealing with all that later in the semester, though, so don't worry about it yet. For now, all you need to remember is that g is 9.8 m/s2 at the surface of the Earth, directed down.
A typical one-dimensional free fall question (free fall meaning that the only acceleration we have to worry about is g) might go like this.
You throw a ball straight up. It leaves your hand at 12.0 m/s.
(a) How high does it go?
(b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight?
(c) How fast is it travelling when you catch it?
Origin = height at which it leaves your hand
Positive direction = up
(a) At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation:
This gives:
0 = 144 + 2 (-9.8) x
Solving for x gives x = 7.35 m, so the ball goes 7.35 m high.
(b) To analyze the rest of the problem, it's helpful to remember that the down half of the trip is a mirror image of the up half. In other words, if, while going up, the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its maximum height, and then double taht to get the total time.
Another way to do it is simply to plug x = 0 into the equation:
This gives 0 = 10 t - 4.9 t^2
A factor of t can be cancelled out of both terms, leaving:
0 = 10 - 4.9 t, which gives a time of t = 10 / 4.9 = 2.04 s.
(c) The answer for part (c) has to be 10 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation:
v = vo + a t
which gives:
v = 10 - 9.8 (2.04) = -10 m/s.
