In physics, we are concerned both with objects at rest (static) and objects in motion (dynamic). Speed is an important measure of the quantity and character of motion. In math classes, we often refer to speed as rate, or rate of travel. Speed is defined as the distance traveled divided by the time it takes
In what follows, we'll be working on linear speed, speed of an object traveling in a straight line, not along a curve.
$$\text{speed} = \frac{\text{distance}}{\text{time}}$$
It's worth pausing here to note that physics is full of three-variable equations like this, and much of the work you'll do to model and solve physical problems will consist of rearranging and solving equations like this with simple algebra. Here are the three rearrangements of the speed equation and how they're obtained:
There's a useful trick to these kinds of rearrangements if your algebra is weak or rusty, the
If you drive from Salida, CO to Denver, CO, you'll cover a distance of about 150 miles (240 Km), and it will take you about 3 hours. What you can calculate from that is average speed:
$$s = \frac{d}{t} = \frac{150\:mi.}{3\:h} = \bf 50 \frac{mi.}{h}$$
Pro tip: in science, we don't really use the common abbreviation "mph" for "miles per hour." We use the fraction
$$\frac{mi.}{h}$$
It's important to notice that what we've actually calculated is an average speed. In fact, on that trip, there are several stop signs, mountain passes and other places where speed can be faster or slower than the average. We don't know anything about any of those details when we calculate average speed, just the total time and distance.
Instantaneous speed, the speed of an object, like a car, at any one instant in time is harder to calculate (it requires calculus – see the box below), but we can measure instantaneous speed at any time in a car by looking at the "speed-o-meter."
Often we put a bar over a quantity to denote that it is an average, so average speed is sometimes written as $\bar s$.
It takes 1:50 (one hour and 50 minutes) to cycle from Sharon, MA to Providence, RI, a distance of 29.3 miles. Calculate the average speed of such a trip in mi./h
Now the speed average speed is just:
$$ \begin{align} s &= \frac{d}{t} = \frac{29.3\:mi.}{1.833\:h} \\[5pt] &= \bf15.9 \: \frac{mi.}{h} \end{align}$$
We should probably round that to 15.9 or 16 mi./h because we only know the distance to the tenths place – it's just good practice.
Note: Once in a while you'll see mi./h written as mi.h^{-1}. (recall that x^{-1} = 1/x) It's written that way because printed text like what you're reading now looks best on one line, and because it's poor practice to write fractions with a slash / on the same line – it's a good way to lose track of units.
It's a common practice to write fractions on one line with a slash for a fraction bar, like $a/b$. That's fine, and really convenient for typesetting (typing all on one line). But it can lead to no good if you do it in calculations, so it's really best avoided.
Always write fractions in calculations vertically, like this:
$$\frac{2a}{b} \; \; \text{ or } \; \; \frac{m}{s} \; \; \text{ or } \; \; \frac{Kg \cdot m}{s^2}$$
The problem with the slash is that sometimes you'll forget what's really in the denominator and that can mess up your calculation.
A coast-to-coast trip across the United States is one of roughly 3000 miles. If a cyclist can maintain an average speed of 17 mi./h on such a trip, how long should the trip take?
$$s = \frac{d}{t}$$
To solve for time, first multiply both sides by time. A good rule to follow is: If the variable for which you want to solve is in the denominator, first ... get it out of there.
$$ts = d$$
Now divide both sides by the speed to get:
$$t = \frac{d}{s}$$
Now we can plug in distance and average speed to get the time. Pay attention to the units. They can tell you when you're messing up.
$$t = \frac{3000\:mi.}{17\frac{mi.}{h}}$$
Notice that dividing by mi./h is the same as multiplying by the reciprocal, so the units of miles divide away nicely
$$t = \frac{3000\:mi.}{17}\:\frac{h}{mi.}$$
The resulting time in hours is:
$$\bf t = 176\:hours$$
That's about 22 8-hour days.
A spacecraft reaches its destination by slowly accelerating (increasing its speed) throughout the whole trip. It begins at a speed of zero m/s and arrives at a speed of 6,000 m/s. If the whole trip takes one year, how long will the craft have traveled?
$$s = \frac{0 + 6000}{2} = 3000\:\frac{m}{s}$$
Now we need to get the times on the same scale. Let's use seconds, so we'll have to convert years to seconds:
$$ \require {cancel} \begin{align} 1\cancel{y} &\left[\frac{365 \cancel{d}}{1\cancel{y}}\right]\left[\frac{24\cancel{h}}{1\cancel{d}}\right]\left[\frac{60\cancel{min.}}{1\cancel{h}}\right]\left[\frac{60 \,s}{1\cancel{min.}}\right] \\[5pt] &= 31,536,000\,s \\[5pt] &= 3.15 \times 10^6\,s \end{align}$$
It's usually better to use scientific notation when numbers get so big. Now the definition of speed is:
$$s = \frac{d}{t}$$
We rearrange before plugging in numbers. Solving for distance gives us:
$$d = s·t$$
Now plugging in numbers and canceling units gives the distance.
$$ \require{ cancel } \begin{align} d &= 3000 \frac{m}{\cancel{s}}\,(3.15 \times 10^6 \cancel{s}) \\[5pt] &= 9.45 \times 10^9 \,m \\[5pt] &= \bf 9.45 \times 10^6\,Km \end{align}$$
That's a lot of Kilometers.
1. | It takes 32 minutes to drive 26.5 miles from Sharon, MA to Providence, RI. Calculate the average speed of the trip in miles per hour. | |
2. | How much longer does it take to drive 26 miles at 65 miles per hour than at 70 miles per hour? | |
3. | A cyclist pedals for 2 hours and 45 minutes and her cycle computer shows an average speed of 18 miles per hour. How far did she pedal? |
When distance traveled is plotted vs. elapsed time, the slope of the graph (rise over run) is the average speed.
Often in math we get the impression that units and even the relative scales of the y and x axes (distance and time here) must be the same. It isn't so. In this graph, distance in meters divided by time in seconds gives speed in m/s, an acceptable unit.
The graphs below illustrate a few different kinds of d vs. t graphs you might encounter. You should learn to interpret graphs like these.
Here are a few examples of position (along the x-coordinate) vs. time (t) graphs for you to practice interpreting. Explanations for each chunk of time are given below each graph, and you can
1. Between t=0 and t_{1} the object is moving in the +x direction at a constant speed, given by the slope of the line, $\Delta x / \Delta t.$ Between t_{1} and t_{2} the particle is not moving, and remains at position x_{2}. The slope of the horizontal line across this region is zero, so the speed is zero. Between t_{2} and t_{3}, the object travels at constant velocity in the -x direction. Finally, after t_{3}, the object is moving in the +x direction, but not at constant velocity. The slope of the curved section decreases over time, so the particle is slowing down as it moves.
2. Between t=0 and t_{1}, the object moves at constant speed in the +x direction. At t_{1} it turns abruptly and heads in the -x direction at constant speed. Between t_{2} and t_{3} the object is stationary because the x vs. t graph has zero slope. After t_{3} the object again moves in the +x direction, but its speed in that direction increases over time. We know this because the steepness of the curved section increases (it gets steeper) as t grows
3. The object remains at rest (speed = 0) at x_{1} from time t=0 to t_{1}. It then moves back to the origin at constant speed. Between t_{2} and t_{3}, the object moves at constant speed in the +x direction until it reaches x_{3}. There it turns in the opposite direction and moves with constant speed until it reaches x_{2} at time t_{4}. From there on it remains stationary.
4. The object starts at rest (speed = 0) until t_{1}, then it moves at constant speed until it reaches position x_{1} at t_{2}. Between t_{2} and t_{3}, the object moves at variable speeds (it accelerates and decelerates. Its speed is positive and decreasign until t_{3}, then negative and increasing until t_{4}. It then travels in the -x direction at constant speed.
4. | If a race car completes a 3-mile oval lap in 58 seconds, calculate its average speed in mi./h. Did the car accelerate during the 58 seconds? | |
5. | The Moon orbits Earth in a near-circular (elliptical) orbit of radius 3.84 x 10^{8} m in 27.3 days. Calculate the average speed of the moon in its orbit. | |
6. | The diameter of Earth is about 12,800 Km. It takes Earth 24 h to rotate once on its axis. Calculate the average speed of any object on the surface of the Earth at the equator. Convert to mi./h, too. | |
7. | The typical airborne speed of an intercontinental B747 jet is 530 mi./h, while the speed of the supersonic (faster than sound) Concorde was 1500 mi./h. If each airliner were to circumnavigate Earth (25,000 mi.) what would be the difference in air time spent by the two aircraft? | |
8. | It takes light (speed = 3.0 x 10^{8} m·s^{-1}) 8.33 min. to travel from the Sun to Earth, and 1.3 seconds from the moon to Earth. Calculate the average distances of the Sun and Moon from Earth. |
Often when we're working with speeds, we're working with relative speeds, or the speeds (or rates) of two things that are linked. There are some nifty techniques for dealing with problems of this kind.
Here are a couple of examples of such problems. They make good use of your organizational and algebra skills.
Runner A can run 1600 m in 6.3 min. Runner B can run the same distance in 7.2 min. How much of a head start, in meters, must runner A give runner B in order for them to finish a 10,000 m race in a tie?
The speed of runner A is
$$ \require{cancel} \begin{align} s_A &= \frac{d}{t} = \left( \frac{1600\,m}{6.3\cancel{min}} \right) \left( \frac{1\cancel{min}}{60\,s} \right) \\[5pt] &= 4.2328\, \frac{m}{s} \end{align}$$
and the speed of runner B is
$$ \require{cancel} \begin{align} s_B &= \frac{d}{t} = \left( \frac{1600\,m}{7.2\cancel{min}} \right) \left( \frac{1\cancel{min}}{60\,s} \right) \\[5pt] &= 3.7037\, \frac{m}{s} \end{align}$$
I've converted these speeds to meters/second because it's a very common unit, but that's not absolutely necessary.
Now we can rearrange the speed equation to solve for time:
The time it takes for the faster runner to cover 10,000 meters is
$$t = \frac{d}{s} = \frac{10,000 m}{4.2328 \frac{m}{s}} = 2362.5\,s $$
That's just a little over 39 min. In that time, runner B will cover a shorter distance. Rearranging the speed equation again to d = st, we get:
$$ \require{cancel} \begin{align} d = st &= 3.7037 \frac{m}{\cancel{s}} \cdot 2362.5\cancel{s} \\[5pt] &= 8,750\,m \end{align}$$
So in order to finish in a tie, runner A will have to let runner B have a 10,000 - 8,750 = 1,250 m (or 1.25 Km) head start.
Runner A and runner B start on a 400 m track at the same time. Runner A laps runner B after 6.5 laps. If they are to run a 5,000 m race, how much time should A wait in order for them to finish in a tie?
We take advantage of the same time on the track run by calculating the speeds each runner:
Solving for time, t = d/s, we can eliminate the time to get:
$$\frac{2600}{s_A} = \frac{2200}{s_B}$$
Cross multiplication gives us
$$2600\,s_B = 2200\,s_A$$
and solving for s_{B} tells us that the speed of runner B is 11/13 the speed of runner A. We don't know the actual speeds, but we do know the relationship between them.
$$s_B = \frac{2200}{2600}\,s_A = \frac{11}{13}\,s_A$$
and
Now the graph of distance vs. time is a line, for which the slope is the speed. From this we can infer a couple of facts:
and
Finally, the difference is 2/13, so we have the head-start distance:
$$\frac{2}{13} (5000\:mi.) = 769\:mi.$$
This section is entirely optional, but read on if you know something about calculus and you'd like to learn about its role in physics. By the way, if you don't know anything about it, don't worry. Calculus really isn't a huge mystery, and it will make perfect sense when you do learn it.
Because the first derivative of a function is its slope, the first derivative of the position (x) with respect to time is the speed.
$$s = \frac{d\,x(t)}{dt}$$
Now if we view the integral as a sum, then the net distance traveled, provided we have some functional form for the speed, is just the integral of that function between two times, t_{1} and t_{2}
$$x = \int_{t_1}^{t_2}\:s(t)\,dt$$
There are many other derivative and integral relationships in physics. Any rate is the derivative of some underlying function, and any time a quantity that can be represented as a continuous function is summed – that's an integral.
Here are eight examples of simple mechanics problems involving speed, distance, time and acceleration (acceleration is covered in another section).
Calculate the linear speed of the moon as it orbits Earth.
Calculate the speed of a person standing "still" at the equator of Earth. It's fast!
Two runners begin a race; one laps the other. From the data, determine how much of a head start the slower runner would need to tie a 5 Km race.
A train passes a person on a station platform. From the speed of the train and the time it takes to pass the person and the platform, determine the length of the platform.
Calculate the initial velocity of an accelerating car, given the acceleration, the distance traveled while accelerating & the time of acceleration.
A ball rolls down a ramp with constant acceleration. Calculate the speed and distance traveled after 3s, and determine distance rolled given a velocity.
Calculate the acceleration experienced by a driver crashing a car into a tree, and convert it to "g-force."
Minutes of your life: 3:35
A ball is launched straight upward at a known initial velociy. Calculate how high it will rise and how long the round trip will take.
Minutes of your life: 2:56
The transitive property of algebra says that if two things are equivalent to the same thing, then they are equivalent to each other.
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