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### Course: MCAT > Unit 9

Lesson 2: Translational motion and calculations- Speed and velocity questions
- Acceleration questions
- Calculating average speed and velocity edited
- Solving for time
- Displacement from time and velocity example
- Instantaneous speed and velocity
- Acceleration: At a glance
- Acceleration
- Airbus A380 take-off time
- Airbus A380 take-off distance
- Why distance is area under velocity-time line
- Average velocity for constant acceleration

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# Acceleration: At a glance

## What is acceleration?

Imagine you are playing golf, and you’ve just set up the golf ball on the tee. You test your grip, take a few practice swings, and peer into the distance at the flag pole marking the first hole for the course. Your wise golf ball guru (who happens to also teach physics) tells you to “Be the ball, young grasshopper.” So you close your eyes and picture the distance the ball must travel. “Yes, young earthworm”, your sensei says, “that distance you see is the displacement you must achieve, displacement is the change in position from one place to another.”

You meditate on this nugget of wisdom, and you start to feel like you and the golf ball are one. But the hole still seems so far away: “How do I change the position of the ball, sensei?

Your sensei replies: “Ahh, the ball cannot simply teleport instantly, it requires velocity; you must change the position of the ball over time.”

In a brilliant burst of enlightenment, you know what you must do. The golf ball sits there minding its own business; with a velocity of 0, and it doesn’t have any plans on moving anytime soon until… WHACK! You hit that golf ball with your club and the ball goes flying through the air. Suddenly the ball is moving, and fast! You know you’ve made your sensei proud, the ball is being displaced, sailing from one position to another toward the hole, This change in position as time passes is what we call velocity.

To change your velocity you must have acceleration. Acceleration is the change of velocity as time passes -- anytime we speed up or slow down, or change direction, we undergo acceleration.

Let’s say we measure the velocity of the golf ball twice. Once just before we hit the ball, and once right after. We might get the following results:

Velocity of golf ball before being hit: 0 meters / second (m/s)

Velocity of golf ball 0.2 seconds after being hit: 15 meters / second (m/s)

Velocity of golf ball 0.2 seconds after being hit: 15 meters / second (m/s)

We can describe acceleration as the $a=\Delta v/\Delta t$ to represent this relationship where ‘a’ is the average acceleration, ‘v’ is velocity, and ‘t’ is time. The Δ is a Greek symbol that means “change”.

*change in velocity*over time, and we can use the shorthand equationIt also turns out we can actually represent velocity in similar symbolic terms, because average velocity is defined as a $v=\Delta d/\Delta t$ . Anytime we have some attribute changing over time, we can call that a ‘rate of change’. So velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

*change in displacement*(d) over time: soNow that we’ve cleared all that up, let’s get back to our golf game. We’ve hit the golf ball, and in 0.2 seconds the ball has changed from not moving at all to moving at 15 m/s. Let’s just plot out that much on a graph.

We could write out our average acceleration as:

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

It’s hard to read 75 (m/s) / second, but because velocity is measured in meters per second (m/s), and acceleration is the change in velocity per second ($\Delta v/s$ ), we can shorten this to 75 m/s${}^{\text{2}}$ . We would then call this 75 meters per second squared.

## Relationships between distance, time, acceleration, and velocity

Unfortunately, unless we had a rocket powered golf ball, the ball would not continue accelerating. It would simply move approximately at a constant velocity in the horizontal direction. As it sailed majestically forward, and our graph would become a flat horizontal line (although the air might slow it down ever so slightly.)

Since we want to keep going fast, let’s set aside our golf clubs and switch gears and to think about a car accelerating forward. Even though a car is much larger, the principle of acceleration is the same!

Suppose we have a car that starts off not moving, but soon begins to accelerate in a straight line up to 50 m/s. The car drives a certain distance (d) along the red line as it accelerates, and the graph plots the car’s change in speed during the trip.

On the graph itself, the x-axis represents time, and the y-axis represents velocity. We can see that as time progresses, the velocity of the car increases. So it must be accelerating.

You can think of each dot on the line as a snapshot of the velocity for each second, and then this means the slope of this curve ($v/\Delta t$ ) represents our acceleration. A steeper portion of the line would mean the car has a larger acceleration. If the car were to hit the brakes, the line would start to bend

*downward*, and we would say that is has a*negative acceleration*, or*deceleration*. Just because we have negative acceleration does not mean we are traveling backwards! Negative acceleration means we are moving from a faster velocity to a slower one!## Applying formulas to real life

Is there anyway we can figure out how far the car has traveled (its displacement) after those 4 seconds of acceleration at a constant velocity? In other words, can we use our understanding of acceleration to solve for $\Delta d$ ?

It turns out there is a way, given the specific rule that we move at a constant acceleration throughout. If we can follow that rule, then we can find displacement using the formula $d={\displaystyle \frac{1}{2}}a{t}^{2}+{v}_{0}t$ .

This formula explains the relationship of displacement to velocity and acceleration. ‘v${}_{0}$ ’ is a symbol for starting or initial velocity, but since we weren’t moving when we first started accelerating, we can set this to 0 and not have to think about it! So just focus on $d={\displaystyle \frac{1}{2}}a{t}^{2}$ , but don’t forget, if our example involved a car that was already moving, we have to take v${}_{0}$ into account!

What this formula is telling us is that if we know the acceleration of an object, and the length of time the object is accelerating, we can calculate exactly how far that object moved! So let’s use it for our car:

d: displacement (meters)

a: acceleration 12.5 m/s${}^{\text{2}}$

t: time 4 seconds

a: acceleration 12.5 m/s

t: time 4 seconds

we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t

remember that we can set the v${}_{0}$ t portion of the formula to 0. multiplication across gives us 100 as an answer, and since we have m/s${}^{\text{2}}$ multiplied by seconds${}^{\text{2}}$ , the two cancel out, leaving us with just meters as units,

d = 100 meters

So in 4 seconds, accelerating at a rate of 12.5 m/s${}^{\text{2}}$ our car was able to drive 100 meters!

## Complicating our travel plans

Now we certainly aren’t going to be accelerating throughout an entire road trip, we’d end going way too fast! So what if we want to consider a scenario where the car is maintaining speed for a short period of time, and then slows down. Maybe instead of accelerating at a constant 12.5m/s${}^{\text{2}}$ , our car speeds up to 40 m/s in 3 seconds, and then it maintains that velocity for 3 more seconds, before hitting the brakes and slowing down to 0 over 4 seconds.

It is pretty tough to tease out all the different rates of acceleration in an example like this, so this is where a graph of our velocities over time would be really handy. What might that graph look like you ask? Well it’s good you asked!

It turns out that instead of having to use our formula three separate times for each change in velocity, we can actually find the area underneath the line to figure out how far we’ve gone!

Using the graph, we can separate the area into two triangles and a rectangle. Then adding up the three areas can give us the distance traveled by our car.

d = (½ base x height) + (base x height) + (½ base x height)

d = ½ (3 sec x 40 m/s) + (3 sec x 40 m/s) + ½ (4 sec x 40 m/s)

d = 60 m + 120 m + 80 m= 260 m

d = ½ (3 sec x 40 m/s) + (3 sec x 40 m/s) + ½ (4 sec x 40 m/s)

d = 60 m + 120 m + 80 m= 260 m

So our distance travelled after 10 seconds is 260m.

## Geometry as a tape measure

So why does this method work so beautifully? Well we know that acceleration is the change in velocity over the change in time; but when we are multiplying the base by the height of the graph, we are actually multiplying the velocity by the time. So let’s look at the time from 3 seconds to 6 seconds. From second 3 to second 4, we are moving at 40 m/s, and we’ve moved for 1 second, so we should have moved exactly 40 m. From second 4 to second 5, we’re still moving at 40 m/s, so to find the total distance in those two seconds, we simply add them together. Do this a third time, and you can simply multiply 3 by 40 to find our result. So the area underneath the lines will give us our displacement, and by adding these areas, we can find our total distance traveled!

One implication of this is that an object only has a nonzero velocity when its position versus time graph has a nonzero slope. Likewise, an object only has acceleration when its velocity versus time graph is sloped.

## Consider the following… jerk

Well now that we know about the change of velocity over time, is there change of acceleration over time? Strangely enough, physicists do have an answer to this question: and the answer is “Jerk”. You can think of jerk this way: when you apply the brakes in a car, you will slow down over time, but nearing the end of slowing down, you suddenly press as hard as you can on the brake, the car will jerk to a stop. That sudden increase in how quickly you stop? That is jerk.

So there’s change in velocity over time, which is acceleration, and change in acceleration over time, and that is called jerk. But it turns out we can keep on going with this game: we can even describe jerk over time! Physicists call this “jounce” or “snap” which can be used to describe time-varying forces acting on things that can change direction very fast, like remote control drones. Of course, not to be outdone, physicists have even proposed different semi-serious names for additional rates of change, including: “Crackle and Pop”, and "Lock, Drop, Shot and Put".

## Want to join the conversation?

- The article says "after those 4 seconds of acceleration at a constant velocity" but it should say "after those 4 seconds of constant acceleration".

The article says "we more at a constant acceleration" but it should say "we move at a constant acceleration".(14 votes)- Correct! Kinematics equations only work if acceleration is assumed to be constant......(7 votes)

- This maybe a silly math questions since I'm pretty rusty.

In the section / paragraph "Apply formulas to Real Life"

d = 1/2 x 12.5 m/s^2 x 4s^2, 1/2 x 12.5 = 6.25

6.25 x 4 = 25 m

How did you get 100 m?

Did you get 100 m because you take 12.5 X 2/1 = 25, 25 X 4 = 100 m?

Thanks in advance!(8 votes)- In the equation you wrote in the first part of your question, the 12.5m/s^2 has the seconds part of that squared, not the entire thing (which is the correct way to write it). However, for the 4s^2, it's actually (4s)^2, which would result in 4^2 x s^2, producing 16s^2. There was just an error in the way they wrote the use of the equation by not specifying that the term (4s) was squared, leaving the confusion that it was 4s^2. Using that, the full final equation would read d = 1/2 x (12.5 m/s^2) x (16s^2) which results in d = 100m(3 votes)

- I'm confused as to how the formula is d=1/2at^2 + Vo t

I don't see where the 1/2 comes from

A=(vf-vo)/t

Vf=(d/t)

I get d=at^2+Vot(4 votes)- The reason you have one half is because you're using an average.

v=d/t therefore d=vt

Once you have that the v is average velocity therefore your average includes that of the final and initial velocities to give: 1/2(vi+vf)

d=1/2(vi+vf)(tf-ti)

d=1/2(vi+vf)(tf-0)

d=1/2(vi+vf)t

d=1/2(vo+v)t since v initial is vo and t represents final or total time.(3 votes)

- so..."jerk" is like in a manual shifting car after you shift to higher gear it moves back and then jumps forward quickly?(3 votes)
- Perhaps your example of shifting to a higher gear is "jerk" in the opposite direction of the "jerk" that is experienced when coming to a stop.(2 votes)

- A police car is traveling at a speed of 50 meters per second when it suddenly accelerates at a rate of 5 m/s2. How fast will the car be going (final velocity) if it maintains that acceleration for 30 seconds.(2 votes)
- If I am doing my math correctly, its final velocity should be 200 m/s. The formula I used was

Acceleration(a) = (Final Velocity (Vf)- Initial Velocity(Vi) / Time. Plug the numbers in and it should be 50 = (Vf-50)/30. Now we solve for Vf. First you multiple 50 by 30, so it becomes. 150 = Vf-50. Now we add 50 to 150 and so we get 200 = Vf. Add the in the units and the answer should become 200 m/s after 30 seconds. Hope this helps.(1 vote)

- why does the unit of time enter twice in the unit of acceleration(2 votes)
- change in m/s per second = (m/s)/s = m/s^2(1 vote)

- At 10seconds isn't the height is at 0 in the graph?so how come after 6 secs the height is calculated as 40 m/s2 in the formula since the car is decelerating to 0 in the next 4secs?(1 vote)
- It says a negative deceleration is when the car hits the brakes, then it says when the car is moving at a slower velocity. does that mean the car didn't stop when it hit the brakes?(1 vote)
- A ball is rolling along the floor with a constant velocity of 4 m/s. How far will it have gone after 48 seconds?(1 vote)
- It depends on how good your throw was with a constant velocity of 4m/s.(1 vote)