What is the acceleration of a van that maintains a constant velocity of 80 kph for 30 minutes

Acceleration is any change in the velocity vector

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 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,

vi, vf = initial & final velocities
ti, tf = initial & final times

When working with linear acceleration, we often just use an abbreviated expression,

$$a = \frac{v}{t}$$

A note on Delta (Δ) notation

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."

Units of acceleration

Acceleration is velocity divided by time, so its units are $\frac{m}{s^2}$ or $m \cdot s^{-2}$ (recall that a negative exponent means "take the reciprocal", so $s^{-1} = \frac{1}{s}$ and $s^{-2} = \frac{1}{s^2}$):

$$\frac{\color{#E90F89}{\text{velocity}}}{\text{time}} = \frac{\color{#E90F89}{\frac{m}{s}}}{s} = \frac{m}{s} \cdot \frac{1}{s} = \bf{\frac{m}{s^2}}$$

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/s2 or m·s-2.

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/s2

(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?

Solution (a) The key to solving this problem is the definition of acceleration:

$$a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}$$

where we know the acceleration (3.5 m/s2) , the change in time (3 s), and the initial velocity, vi = 0. I always like to rearrange the equation first before I plug in numbers:

Now the final velocity is

$$v_f = 3.5\,\frac{m}{s^2} · 3\,s = 10.5\,\frac{m}{s}$$

Solution (b) Generally if we're looking for a distance, we use an average velocity. With a constant acceleration, the average velocity is just the average of the initial and final velocities (add them up and divide by 2), and in this case, vi = 0, so the average velocity is just 10.5/2 m/s = 5.25 m/s. Now we just rearrange the definition of velocity so that we can solve for distance (x),

$$v = \frac{\Delta x}{\Delta t} \: \longrightarrow \: x = v \Delta t$$

so the distance is

$$ \require{cancel} x = 5.25 \, \frac{m}{\cancel{s}} \cdot 3 \cancel{s} = 15.75 \, m$$

Solution (c) This time we have the final velocity, vf = 15 m/s, so our average velocity will be 15/2 = 7.5 m/s, but we don't have the time over which this occurs. We'll have to go back to the definition of acceleration for that:

Now rearrange to find time:

$$\Delta t = \frac{v_f - v_i}{a}$$

And plug in what we know to get the time:

$$ \Delta t = \frac{15\cancel{m}·s^{-1}}{3.5\cancel{m}·s^{-2}} = 4.285\,s$$

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:

$$ x = v \Delta t = 7.5\,\frac{m}{\cancel{s}}\,(4.285\cancel{s}) = \bf 32\,m$$

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.

Solution: There are a couple of important things to realize about a problem like this. First, once a projectile (the ball) is released from whatever force launches it, there is no more upward acceleration at all; any remaining acceleration is the downward acceleration of gravity, which we call g (g = 9.8 m/s2). That downward acceleration is slowing the ball on its way up and speeding it up on its way down.

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/s2 downward, so we can write an acceleration equation:

The only unknown (thanks to the fact that vf = 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:

$$ \require{cancel} \begin{align} \Delta t = \frac{v_f - v_i}{a} &= \frac{-27\cancel{m}·s^{-1}}{9.8\cancel{m}·s^{-2}} \\[5pt] &= 2.755 s \end{align}$$

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

$$ \require{cancel} \begin{align} d = vt &= 13.5\,\frac{m}{\cancel{s}}(2.755\cancel{s}) \\[5pt] &= 37.2\,m \end{align}$$

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 vi = 0, so the distance traveled is

$$d = vt = \frac{v_f · t}{2}$$

where vf/2 is just the average velocity when vi = 0, and vf isn't yet known. We can also write the acceleration for this leg (again because vi = 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 vf like this:

$$v_f = \frac{2d}{t} \: \: \color{#E90F89}{\text{ and }} \: \: v_f = at$$

We can put them together, eliminating vf (and the need to know it), and with a little rearrangement, we get:

$$\frac{2d}{t} = at \: \color{#E90F89}{\longrightarrow} \: d = \frac{1}{2} at^2$$

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:

$$d = \frac{1}{2} at^2 \: \color{#E90F89}{\longrightarrow} \: t = \sqrt{\frac{2d}{a}}$$

$$ t = \sqrt{\frac{2 \cdot 37.2\cancel{m}}{9.8\cancel{m} \cdot s^{-2}}} = \bf 2.755\,s$$

Notice that it's exactly the same time as the upward journey took, which means that

• the average velocity on the downward leg will be the same as that of the upward leg
• the distance traveled will be the same and
• the final velocity at the bottom of the trip will equal the initial upward velocity; it will just be reversed in sign.

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.

Time up = Time down

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.

A box, initially at rest, slides down an inclined plane with a uniform acceleration and attains a velocity of 27 m/s in 3 seconds from rest. Calcluate the final velocity and distance moved in 6 seconds.

Solution: We have all of the information we need in this problem to find the acceleration of the box:

Now with that acceleration we can find the final velocity of the box after 6 seconds:

We can rearrange to calculate vf , to get:

$$v_f = a·\Delta t = 9\,\frac{m}{s^{\cancel{2}}}(6\,\cancel{s}) = 54\,\frac{m}{s}$$

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

$$d = vt = 27\,\frac{m}{\cancel{s}}(6\,\cancel{s}) = 162\,m$$

Notice that we could have calculated the distance using the freefall equation (which really should be called the "smooth acceleration from v = 0" equation):

$$\begin{align} d = \frac{1}{2} at^2 &= \frac{1}{2}·9\,\frac{m}{\cancel{s^2}}(36\,\cancel{s^2}) \\[5pt] &= \bf 162\,m \end{align}$$

Cool!

Acceleration is the slope of a velocity vs. time graph, $\frac{\Delta v}{\Delta t}.$ Here are a few velocity vs. time graphs to illustrate some of the features you might encounter. Roll-over / tap each graph to see an explanation of each time region.

The object moves at constant velocity (no acceleration) between time t=0 and t1.

It accelerates in the negative direction (decelerates) to zero velocity by time t2.

The object accelerates with constant acceleration (the graph is a line with constant slope) between t2 and t3.

Between t3 and t4, the object has constant negative acceleration (decelerates) until the velocity = v2.

From t4 on, the object has zero acceleration, and moves at constant velocity, v2.

Between t=0 and t1 the object accelerates with constant acceleration (because the graph is a line with constant slope) up to a velocity of v2.

Between t1 and t2 the object does not accelerate because the slope of the v vs. t graph in that region is zero. It moves with constant velocity, v2.

The object decelerates (constant negative acceleration or slowing down) to velocity v = v1.

After t3, the particle accelerates in the forward direction, but the magnitude of that acceleration changes: the rate of acceleration gets smaller as time passes. In physics, this change in acceleration is called "jerk."

Up to time t1, the object depicted in this graph is neither moving nor accelerating (if it's not moving, it can't be accelerating).

It accelerates in the positive direction to velocity v2 by time t2. The acceleration is constant because the graph is linear in this time period.

Between t2 and t4 the object accelerates to a peak velocity of v3, then decelerates back to v2. Both the acceleration and deceleration are non-linear. That is, they both change over time. When acceleration changes over time, that change, (Δ a) is called "jerk.""

After t4 the object decelerates at a constant rate.

Circular Motion: Centripetal Acceleration

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.

Acceleration and calculus

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(to) 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