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Kinematics

Written by tutor Scott S.

Kinematics in 1 dimension, using algebra

At the 2011 Berlin Marathon, runner Patrick Makau of Kenya ran a mile in four minutes, forty-three seconds – and maintained that 4:43 mile pace for 26 miles, 385 yards (26.2 miles, or 42.2 kilometers), to finish the marathon in a world-record time of 2:03:38 (two hours, three minutes, thirty-eight seconds). Yet, a determined spider could have crawled fast enough to crawl across Patrick’s shoe as the starting gun fired, and then crawled across his shoe again as he crossed the finish line.

How can this be, since a typical spider only walks a few inches each second? The answer, of course, is that the spider could have traveled in a straight line, whereas the course that Patrick ran was full of curves. (You can view the Berlin marathon course here: http://www.bmw-berlin-marathon.com/en/event/course.html) In order to talk about ideas such as “who covered more ground,” we first have to distinguish between some similar but confusing terms.

Vectors and Scalars

Sometimes, we care about direction, and sometimes we don’t. A vector is any quantity where direction matters as well as amount, and a scalar is any quantity where direction doesn’t matter, only the amount does. (The amount, or number and unit, associated with a vector, is called its magnitude.) A physics vector “carries” something from one place to another in the same way that a disease vector, like a mosquito, carries diseases. The word scalar, which rhymes with “sailor,” reminds me of a bathroom scale, which gives me a number, but no direction. And “magnitude” sounds like “magnify,” so it should make sense that the magnitude of a vector is just its size.

Suppose your friend Arnold says “I’m so fit I could swim five hundred meters.” It doesn’t matter which direction he swims, so “500 meters” is a scalar. On the other hand, what if he radios the lifeguard “Come rescue me, my boat’s sinking and it’s too far to swim to shore! I’m 500 meters northeast of you”? Is it important that the lifeguard take his motorboat in the right direction? Arnold certainly thinks so! Therefore, “500 hundred meters northeast” is an example of a vector: it has magnitude (500 meters) and direction (northeast).

Alternatively, Arnold could have specified his position as “354 meters north and 354 meters east of you.” That would be another way of describing the same vector. We’ll touch more on this later. For more, see the WyzAnt page on vectors.

Distance traveled and displacement

Distance traveled (scalar): What your odometer would read.
Displacement (vector): the distance from something’s starting location to its final location, coupled with the direction from start to finish. (The verb “displace” means to remove from one place and put in another.)

Let’s find the distance traveled, the displacement, and the magnitude of the displacement for Patrick in the 2011 marathon.

Distance traveled: this is just the distance of the marathon he ran: 26.2 miles, or 42.2 kilometers.
Displacement: this is the distance from the starting line to the finish line, coupled with direction: about 0.5 miles east, or 800 meters east.

Magnitude of displacement: chop direction off of displacement, and you have magnitude of displacement: the magnitude of Patrick’s displacement was 0.5 miles, or 800 meters. (He ended up 800 m from where he started.)

You try it

Find the distance traveled, displacement, and magnitude of displacement for the spider. Express your answers in meters. Type all three answers into the box below, separated by commas

Here, both the distance traveled and the magnitude are 800 m, while the displacement is 800 m east.

{800 m, 800 m east, 800 m|800m, 800m east, 800m|800 meters, 800 meters east, 800 meters|800 m, 800 m E, 800 m|800m, 800m E, 800m}

Are you wondering why distance traveled equals displacement for the spider, but not for the human? It’s because the spider traveled in a straight line, while the human made turns. In general, distance traveled = magnitude displacement for if you go in a straight line and don’t reverse course. If you switch directions, either by curving or by going backwards, then distance traveled > magnitude of displacement.

Speed and velocity

Speed (scalar): distance traveled/time taken
Velocity (vector): displacement/time taken
Here are two ways to calculate Patrick’s speed:
(26.2 miles)/(2 hours, 3 minutes, 38 seconds) = 26.2 miles/2.06 hours = 12.7 miles per hour.
(42.2 km)/(2 hours, 3 minutes, 38 seconds) = 42,200 m/ 7418 seconds = 5.69 meters per second.

You try it

Calculate Patrick’s velocity, and the magnitude of his velocity. Express your answers in meters per second. Type both answers into the box below, separated by commas.

Patrick's velocity needs to have a direction attached to it, because it is a vector. The magnitude of Patrick's velocity does not need to have a direction attached, because magnitude indicates the numerical amount.

{0.108 m/s east, 0.108 m/s|.108 m/s east, .108 m/s|0.108m/s east, 0.108m/s|0.108 m/s E, 0.108 m/s|.108 m/s E, .108 m/s}
Now, calculate the spider’s speed, velocity, and the magnitude of his velocity. Express your answers in meters per second. Type all three answers into the box below, separated by commas.

The spider's speed is a scalar measurement, which does not need a direction. The spider's velocity is a vector, which means it needs to be written with a direction. The magnitude of the spider's velocity, in this case, is the same as its speed, which does not need a direction attached.

{0.108 m/s, 0.108 m/s east, 0.108 m/s|.108 m/s, .108 m/s east, .108 m/s|0.108m/s, 0.108m/s east, 0.108m/s|0.108 m/s, 0.108 m/s E, 0.108 m/s|.108 m/s, .108 m/s E, .108 m/s}
Is Patrick’s speed the same as the magnitude of his velocity? Is the spider’s speed the same as the magnitude of its velocity? Answer each question with a "yes" or "no," separated by commas, in the box below.

Patrick's speed was 5.69 m/s, but the magnitude of his velocity was only .108 m/s. This is because of the change in direction. The spider's speed is the same as the magnitude of its velocity because there was no change in direction.

{no, yes|no,yes|No, Yes}

In general, speed = magnitude of velocity whenever distance traveled = magnitude of displacement: they’re equal if and only if travel is in a constant direction. If you ever turn left, turn right, or make a U-turn, then speed > magnitude of velocity.

Acceleration

Up until now, we haven’t worried about changing velocities or speeds; we’ve just assumed that they’re constant. What if they do change? Speed and velocity tell us how fast an object’s location is changing. Acceleration tells us how fast an object’s velocity is changing.

Acceleration (vector): change in velocity / time taken
Or: acceleration = (final velocity – initial velocity)/(final time – initial time)
The units of acceleration could be (miles per hour) per second, but in physics, we usually use the unit (meter per second) per second, or meter per second squared. Algebraically, (m/s)/s = m/s2.
(In regular English, people use the word accelerate to mean “speed up,” and decelerate to mean “slow down.” But in physics, accelerate means “speed up,” “slow down,” or “change direction.” After all, a change in velocity could be a change in speed or a change in direction.)

Let’s suppose Patrick was standing still when the gun went off, and it took him five seconds to reach his cruising speed of 5.69 m/s. One second after the gun went off, he had only sped up to 2.30 m/s. The beginning of the race had runners heading west. What was his acceleration?

It helps to write out the initial and final velocities and times.
v1 = 2.29 m/s west
v2 = 5.69 m/s west
t1 = 1 s
t2 = 4 s

So a = (5.69 m/s west – 2.30 m/s west)/(4 s – 1 s)
a = (3.39 m/s west)/(3 s)
a = 1.13 m/s2 west.

Kinematics in 2 or 3 dimensions, using trigonometry

Peyton and Demaryius are playing a pickup game of touch football. Demaryius, the wide receiver, lines up right next to Peyton, the quarterback. Demaryius runs 4 yards downfield, cuts to the left, and runs three yards toward the sideline before catching the pass. He is immediately touched, ending the play. The WR’s distance traveled is 7 yards (4 + 3), but his displacement is 5 yards (draw a triangle and use the Pythagorean theorem to find the length of the hypotenuse). How many yards does his team gain on the play?

“It’s a 4-yard gain,” anyone who knows the rules of American football will say. Why four? The WR runs seven yards; the QB throws the ball five yards. Where do we get four? “It’s four yards because he only makes it four yards closer to the goal line. We only care about travel down the field, not toward the sideways.” Aha! Football fan, you have created a coordinate system!

Physicists do the same thing: imagine laying the axes for a graph on the football field, with the y- axis pointing downfield and the x-axis pointing toward the sideline. Then “downfield” becomes “in the y-direction,” and “crossfield” becomes “in the x-direction.” Thus, a physicist would describe the play this way: Δx = -3 yd (the symbol Δ means “change in.”)

Δy = +4 yd

The values Δx and Δy aren’t vectors or vector magnitudes: after all, the magnitude of a vector can’t be negative, like Δx is. Instead, Δx and Δy are called the “x component” and “y component” of the displacement vector.

We can do the same thing in three dimensions. Often, a coordinate system will be set up on Earth’s surface so that the x-axis points east, the y-axis points north, and the z-axis points up. Imagine a buried treasure located 5 meters east of us, 2 meters south of us, and 4 meters underground. Then a vector pointing from us to the treasure has the components:

Δx = 5 m
Δy = -2 m
Δz = -4 m

Keep in mind, we don’t have to orient our coordinate system this way: we’re free to point our axes in whichever direction makes our problem the easiest to solve. Also remember that any vector can be split into components, not just displacement.

Most physics problems in high school and the beginning of college assume that the acceleration in a problem is constant, but velocity might not be. In principle, you only need two equations to solve these problems: the definition of velocity (combined with the ability to compute an average) and the definition of acceleration.

Definition of average velocity: vaverage = d/t and vaverage = (v1 + v2)/2, the arithmetic mean of v1 and v2.

(v1 + v2)/2 = d/t
(v2 + v2)*t/2 = d

Definition of acceleration: a = (v2 – v1)/t
a*t = v2 - v1
v1 + a*t = v2

By solving (1) for a different variable, and plugging the answer into (2), we can generate new equations. These new equations help save us a step when solving problems.

In general, you have five different quantities to worry about: t, d, v1, v2, and a. They’ll give you three of them, and ask for a fourth. The fifth quantity –neither given nor asked for – determines which equation you should use.

1) d = (v1 + v2)*t/2               no a
2) v2 = v1 + a*t                     no d
3) d = v1*t + (1/2)a*t2          no v2
4) d = v2*t - (1/2)a*t2           no v1
5) v22 = v12 + 2*a*d             no t

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