You know what acceleration is because you've been in a vehicle that speeds up, slows down or takes a sharp corner. When the car speeds up (accelerates) it pushes against your back as it accelerates you. When the car slows down, your body has forward inertia (tendency to keep moving with the same velocity) and tends to keep going until your seat belt exerts a backward force on you to slow you at the same rate as the car.
And when the car is turned through a tight corner at high speed, you feel like you're being thrown to the outside—pushed toward the outside door. What's really happening is that your body has inertia in the direction you were going before the turn; now the car is accelerating into the center of the turn and pushing on you.
Acceleration can be a tricky thing, but its definition is not:
Acceleration is any change in the velocity vector. It can be a change in the vector length (the speed) or its direction.
Linear acceleration is easy. If the speed of an object traveling in a straight line increases or decreases, then the length of its velocity vector increases or decreases (speed is the magnitude of the velocity vector), therefore we have acceleration. Often we refer to an increase of speed as acceleration or "positive acceleration" and a decrease as deceleration or "negative acceleration". Linear acceleration is
$$a = \frac{\Delta v}{\Delta t} = \frac{f_f - v_i}{t_f - t_i}$$
where Δv is the change in velocity and Δv is the change in time (the length of the time interval of the acceleration). Also,
t_{i}, t_{f} = initial & final times
When working with linear acceleration, we often just use an abbreviated expression,
$$a = \frac{v}{t}$$
The definition of linear acceleration is shown below. In it, we introduce the symbol Δ, the Greek letter Delta, which will mean "change in". We use Δ quite often in science in math to denote change. ΔT could mean change in temperature, Δx could mean change in position, &c. We read Δv as "delta-vee", and it means "subtract the initial value of v from the final value of v."
alpha | Α | α |
beta | Β | β |
gamma | Γ | γ |
delta | Δ | δ |
epsilon | Ε | ε |
zeta | Ζ | ζ |
eta | Η | η |
theta | Θ | θ |
iota | Ι | ι |
kappa | Κ | κ |
lambda | Λ | λ |
mu | Μ | μ |
nu | Ν | ν |
xi | Ξ | ξ |
omicron | Ο | ο |
pi | Π | π |
rho | Ρ | ρ |
sigma | Σ | σ |
tau | Τ | τ |
upsilon | Υ | υ |
phi | Φ | φ |
chi | Χ | χ |
psi | Ψ | ψ |
omega | Ω | ω |
Acceleration is velocity divided by time, so its units are m/s^{2} or m·s^{-2} (recall that a negative exponent means "take the reciprocal", so s^{-1} = 1/s and s^{-2}= 1/s^{2}):
The SI units of acceleration are pronounced "meters per second-squared" or (less commonly) "meters per-second per-second." Recall that SI stands for Système International d'Unités, an internationally agreed-upon set of units for every measurement.
The units of acceleration are m/s^{2} or m·s^{-2}.
SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units
length | meter | (m) |
mass | Kilogram | (Kg) |
time | second | (s) |
force | Newton | N |
energy | Joule | J |
Because acceleration is just the velocity vector v multiplied by a scalar (1/t), acceleration is also a vector, and you should add it to your growing list of vector quantities.
Time is not a vector, even though, in a sense, it has a direction; it just has the same direction, always. Time is a scalar quantity, only its magnitude is important; it has no direction in the sense that a velocity or a force does.
Often my students like to engage me in great battles about whether time is or is not a dimension. It is not. Time is a parameter in any equation in which it appears. Time is not the fourth dimension, or any dimension. What might surprise you is that "spaces" with tens, hundreds, thousands, ... even an infinite number of dimensions "exist" and are used all the time to solve problems in all kinds of math and science.
A ball rolls down an inclined plane with a constant acceleration of 3.5 m/s^{2}
(a) If a stationary ball is released on the ramp, how fast is it traveling after rolling for 3 s?
(b) How far does the ball travel in those 3 seconds?
(c) How far will the ball have traveled by the time its velocity is 15 m/s?
$$a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}$$
where we know the acceleration (3.5 m/s^{2}) , the change in time (3 s), and the initial velocity, v_{i} = 0. I always like to rearrange the equation first before I plug in numbers:
Now the final velocity is
$$v_f = 3.5\,\frac{m}{s^2} · 3\,s = 7\,\frac{m}{s}$$
$$v = \frac{\Delta x}{\Delta t} \: \longrightarrow \: x = v \Delta t$$
so the distance is x = 3.5 m/s · 3 s = 10.5 m.
Now rearrange to find time:
$$\Delta t = \frac{v_f - v_i}{a}$$
And plug in what we know to get the time:
$$\Delta t = \frac{15\,m·s^{-1}}{3.5\,m·s^{-1}} = 4.285\,s$$
I like to keep a few digits past the decimal around in the middle of a calculation rounding only at the end. Otherwise, there can be significant roundoff error by a calculator as you work through a longer calculation.
Now it's just a matter of plugging that time into our velocity equation, using the average velocity of 7.5 m/s, to get the distance:
$$x = v \Delta t = 7.5\,\frac{m}{s}\,(4.285\,s) = \bf 32\,m$$
A 3 Kg ball is launched straight into the air at an initial velocity of 27 m/s. How far will the ball travel upward? How long will it take the ball to return to the point of launch, and what will its velocity be at that point?
The figure below breaks the problem up into two "legs," upward and downward, and organizes what we know and don't know.
Second, the velocity at the very top of the trajectory is zero, just before the ball begins its downward trip. That fact alone is what allows us to be able to solve problems like this.
On the upward leg, the initial velocity is 27 m/s, and the final velocity is 0 m/s (giving us an average velocity of 13.5 m/s). The acceleration is 9.8 m/s^{2} downward, so we can write an acceleration equation:
The only unknown (thanks to the fact that v_{f} = 0) is the time, so we can calculate it. Notice that we wrote the downward acceleration vector as negative. It's not crucial in this case, but it does give us the right sign when we calculate time. Now we can rearrange and to isolate the time:
$$\begin{align}\Delta t = \frac{v_f - v_i}{a} &= \frac{-27\,m·s^{-1}}{9.8\,m·s^{-2}} \\ \\ &= 2.755 s \end{align}$$
Now let's calculate the distance traveled by the ball in its upward path. That's just the average velocity multiplied by the time:
$$v = \frac{d}{t} \: \longrightarrow \: d = vt$$
then
$$\begin{align} d = vt &= 13.5\,\frac{m}{s}(2.755\,s) \\ \\ &= 37.2\,m \end{align}$$
So the ball will rise to a height of 37.2 m before beginning the downward leg.
Now for the downward leg, we need to think a little more deeply. The initial velocity (see the figure) is v_{i} = 0, so the distance traveled is
$$d = vt = \frac{v_f · t}{2}$$
where v_{f}/2 is just the average velocity when v_{i} = 0, and v_{f} isn't yet known. We can also write the acceleration for this leg (again because v_{i} = 0) as
$$a = \frac{v}{t}$$
I'm dropping the delta on the time to simplify things. Now if we solve both of those equations for v_{f} like this:
We can put them together, eliminating v_{f} (and the need to know it), and with a little rearrangement, we get:
That's called the freefall formula. It's usually written with a = g, the acceleration of gravity. You can learn more about it from the freefall page.
We can use a little algebra to rearrange the freefall equation to calculate the time:
$$t = \sqrt{\frac{2·37.2\,m}{9.8\,m·s^{-2}}} = \bf 2.755\,s$$
Notice that it's exactly the same time as the upward journey took, which means that
That is true for any projectile that returns to the point of launch, and it doesn't even matter that it was launched straight up, as you will see when you study 2D projectile motion.
A projectile launched upward will return to its starting height in the same time it took to reach its maximum height, and at the same velocity (with opposite sign) as its initial velocity.
Now with that acceleration we can find the final velocity of the box after 6 seconds:
We can rearrange to calculate v_{f} , to get:
$$v_f = a·\Delta t = 9\,\frac{m}{s^2}(6\,s) = 54\,\frac{m}{s}$$
From that we can calculate the distance traveled by noticing that the average velocity is
$$v_{avg} = \frac{54 + 0}{2} = 27\,\frac{m}{s}$$
so the distance traveled is
$$d = vt = 27\,\frac{m}{s}(6\,s) = 162\,m$$
Notice that we could have calculated the distance using the freefall equation (which really should be called the "smooth acceleration from v = 0" equation):
$$\begin{align}d = \frac{1}{2} at^2 &= \frac{1}{2}·9\,\frac{m}{s^2}(36\,s^2) \\ \\ &= \bf 162\,m \end{align}$$
Cool!
1. | A driver reaches a speed of 40 m/s seven seconds after starting from a stop. Calculate her acceleration in that time. How far did she travel in that seven seconds? If she were to accelerate at this rate for another 90 s, how fast would she be going? | |
2. | If a rocket in space is moving at a constant velocity of 9.8 m/s and then uses its propulsion system to accelerate to 10 m/s during a 3.0 minute burn, calculate the acceleration of the rocket required to achieve that extra velocity. | |
3. | A car, initially traveling at a constant velocity, accelerates at a rate of 1.0 m/s^{2} for a period of 12 s. If the car traveled 190 m during this 12-second period, what was the velocity of the car when it started to accelerate? | |
4. | On a test track, it takes a Toyota Camry (that's an anagram of "my car") 57.7 m to skid to a stop after braking from an initial speed of 60 mi./h. Calculate the average acceleration of the car during the stopping process. | |
5. | The tallest building in the U.S. is the Willis Tower in Chicago, with a height of 443 m (1454 ft.) Ignoring wind resistance, how fast would a penny dropped from the top be moving when it hit the ground? | |
6. | A projectile is launched from ground level with an initial velocity of 32 m/s. Assuming no air resistance (friction), how high will the ball travel before it turns back to Earth? How long will the round-trip take, and how fast will the ball be going when it is half-way back down? |
Any motion in a curved path, even motion with constant speed, is accelerated motion. Remember that any change in the velocity vector is defined as acceleration. That includes the direction of the velocity vector, regardless of whether its magnitude changes.
What keeps a body moving in a circular path is a centripetal ("center-seeking") force, which produces a centripetal acceleration. I will save the mathematical definition of centripetal acceleration for the section on circular motion.
← This animation shows how the velocity vector changes as a rock tied to a string is twirled in a circle. The rock moves in a circle because of the center-directed force of the string (centripetal force). Notice that no pushing force from behind the rock is required. There will be more to say about this when we discuss forces in a later section.
The velocity vector of the rock at any instant in time is in a direction tangent to the circle and in the same plane. The direction of the velocity vector changes continuously as the rock traces out the circle.
On the first round-trip of the animation several velocity vectors are shown, on the second, all but the real-time vector are stripped away. Look at the loop a couple of times until you get the idea: Any kind of curved motion, regardless of whether there is a speed change, is acceleration.
In the section on Newton's laws, you will see that what you actually feel pushing on you when a car takes a sharp turn is not the acceleration, but a product of it, a force. Acceleration always produces a force, and forces, if applied to a movable object, always produce acceleration.
This link between forces and acceleration is the basis for our understanding of the physics of moving things, which is called dynamics.
We said that acceleration is any change in the velocity vector. In one-dimensional systems, acceleration is the change in speed over time, which is certainly a derivative relationship. In multidimensional systems, where velocity has a direction other than right or left, the acceleration vector is just the derivative of the velocity vector with respect to time.
Acceleration is the derivative of velocity with respect to time:
$$a = \frac{dv}{dt}$$
or acceleration is the second derivative of position with respect to time:
$$a = \frac{d^2x}{dt^2}$$
Because acceleration is a derivative, we can also go backward using integration to find velocity. Velocity is the definite integral of acceleration over a definite time interval:
$$v = \int_{t_1}^{t_2} a\,dt$$
For example, if we wanted to calculate a velocity v(t) and we knew v(t_{o}) and a(t), then
$$v(t) = v(t_o) + \int_{t_1}^{t_2} a\, dt$$
Here are eight examples of simple mechanics problems involving speed, distance, time and acceleration.
Calculate the linear speed of the moon as it orbits Earth.
Minutes of your life: 2:34
Calculate the speed of a person standing "still" at the equator of Earth. It's fast!
Minutes of your life: 2:53
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.
Minutes of your life: 4:48
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.
Minutes of your life: 3:45
Calculate the initial velocity of an accelerating car, given the acceleration, the distance traveled while accelerating & the time of acceleration.
Minutes of your life: 3:43
A ball rolls down a ramp with constant acceleration. Calculate the speed and distance traveled after 3s, and determine distance rolled given a velocity.
Minutes of your life: 3:32
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
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.