somebody explain the idea of acceleration units to me. The whole “seconds squared” or “seconds per second” makes no sense to me.

211 views

Like if a car starts at rest and moves at 4m/s^2 for 10seconds, what does that mean?

Does it mean the car is exponentially increasing in speed? Can someone draw it out for me second by second?

Edit:
***I have a follow up question to several of y’all’s responses in how the concept of acceleration relates to one of the big kinematics equations as well. That’s one of the big discrepancies I’m having trouble understanding***

In: 9

24 Answers

Anonymous 0 Comments

>Like if a car starts at rest and moves at 4m/s2 for 10seconds, what does that mean?

The wording of this is a little nonsensical, so let me start by giving you a new example. Imagine dropping a ball off of the Empire State Building. Gravity pulls things toward Earth, and it turns out it pulls them down at 9.8 m/s^(2). Let’s round that up to 10 for easy maths. At first, you’re holding the ball and aiming. The ball isn’t moving, because you’re holding it. It is moving exactly 0 meters every second, so its speed is 0 m/s. When you let go, the ball is going to start falling towards earth, speeding up as it goes. But how much is it speeding up? This is that 10 m/s^2 bit, exactly 1 second after you drop the ball, gravity will have sped the ball up to 10 m/s. Another second goes by, and the ball will have sped up another 10 m/s, which means it will now be going 20 m/s. At the third second, the ball will have sped up another 10 m/s, for a total of 30m/s. The ball is speeding up by 10 meters per second, every second. Or said another way, the ball is accelerating 10 meters per second per second. Mathematically, both of the seconds end up in the denominator, so some people write it as m/s^2 but it is all the same thing.

>Does it mean the car is exponentially increasing in speed?

No, not necessarily. The example you gave of a car accelerating at 4m/s^2 would be linearly increasing its speed. If acceleration is given as just a number, then it is constant. If that is the case then the body subject to that acceleration will be changing its speed linearly, not exponentially, just like the ball drop example above. It is possible for things to accelerate exponentially, it’s just the value of that acceleration won’t be just a simple number.

>how the concept of acceleration relates to one of the big kinematics equations as well.

The biggest thing to remember is that it all starts with position. In the ball drop example, the ball starts at the very top of the building, which is 381 meters. That’s a position, it is just a number of meters.

From there, we can build on that to get some more information. If I am holding the ball there, the ball isn’t moving, so it’s position isn’t changing. No matter how many seconds I hold that ball, each and every one of those seconds the ball will move exactly zero meters. This means its *velocity* is zero meters per second, or 0 m/s.

*(Velocity is just speed combined with direction. Kinematics cares a lot about velocity, but for conceptual understanding I’m just going to use speed for the time being, even if it isn’t technically correct.)*

So now I know how much its position is changing every second, and I’m calling that speed. But how much is its speed changing every second? Well if it’s speed was 0 m/s before, and a second later it’s speed is still 0 m/s, then its speed is changing exactly 0 m/s every second, or, it’s *acceleration* is 0 m/s^(2).

So let’s say I’m tired of holding this ball, and I drop it. I want to know what its speed will be 3 seconds later. Well, I know the only thing making it move at all is gravity, and gravity is accelerating it at 10 m/s^2 down (like velocity, acceleration is a vector and has direction, and because gravity pulls down we typically say -9.8 m/s^(2), but again for conceptual understanding I’m leaving the signs out of it), so I would use this equation:

vf = vi + (a × t)

vi = 0 m/s; a = 10 m/s^2; and t = 3 s; so plugging that all in:

vf = 0 m/s + (10 m/s^2 × 3 s) = 0 m/s + 30 m/s = 30 m/s

So, that’s one equation, and it tells me the ball will be moving 30 m/s three seconds after I drop it. But let’s just jump straight to the hardest question: how long will it take for the ball to hit the ground? For this I need this equation:

s = (vi × t) + (0.5 × a × t^(2))

s is displacement, and I know the Empire State Building is 381 m tall, so that is how far the ball has to fall to hit the ground. vi is 0 m/s, and anything times zero is zero, so the first term can go away. a is 10 m/s^2 and we want to know t. So, plugging all of that in we get:

381 m = (0 m/s × t) + (0.5 × 10 m/s^2 × t^(2))

381 m = 5 m/s^2 × t^2

t^2 = 381 m ÷ 5 m/s^2 = 76.2 s^2

t = √(76.2 s^(2)) = 8.7 s

So, it will take the ball 8.7 seconds to hit the ground if I dropped it from the top of the Empire State Building.

Hopefully that helps you understand a little better in terms of kinematics.

Anonymous 0 Comments

>Like if a car starts at rest and moves at 4m/s2 for 10seconds, what does that mean?

The wording of this is a little nonsensical, so let me start by giving you a new example. Imagine dropping a ball off of the Empire State Building. Gravity pulls things toward Earth, and it turns out it pulls them down at 9.8 m/s^(2). Let’s round that up to 10 for easy maths. At first, you’re holding the ball and aiming. The ball isn’t moving, because you’re holding it. It is moving exactly 0 meters every second, so its speed is 0 m/s. When you let go, the ball is going to start falling towards earth, speeding up as it goes. But how much is it speeding up? This is that 10 m/s^2 bit, exactly 1 second after you drop the ball, gravity will have sped the ball up to 10 m/s. Another second goes by, and the ball will have sped up another 10 m/s, which means it will now be going 20 m/s. At the third second, the ball will have sped up another 10 m/s, for a total of 30m/s. The ball is speeding up by 10 meters per second, every second. Or said another way, the ball is accelerating 10 meters per second per second. Mathematically, both of the seconds end up in the denominator, so some people write it as m/s^2 but it is all the same thing.

>Does it mean the car is exponentially increasing in speed?

No, not necessarily. The example you gave of a car accelerating at 4m/s^2 would be linearly increasing its speed. If acceleration is given as just a number, then it is constant. If that is the case then the body subject to that acceleration will be changing its speed linearly, not exponentially, just like the ball drop example above. It is possible for things to accelerate exponentially, it’s just the value of that acceleration won’t be just a simple number.

>how the concept of acceleration relates to one of the big kinematics equations as well.

The biggest thing to remember is that it all starts with position. In the ball drop example, the ball starts at the very top of the building, which is 381 meters. That’s a position, it is just a number of meters.

From there, we can build on that to get some more information. If I am holding the ball there, the ball isn’t moving, so it’s position isn’t changing. No matter how many seconds I hold that ball, each and every one of those seconds the ball will move exactly zero meters. This means its *velocity* is zero meters per second, or 0 m/s.

*(Velocity is just speed combined with direction. Kinematics cares a lot about velocity, but for conceptual understanding I’m just going to use speed for the time being, even if it isn’t technically correct.)*

So now I know how much its position is changing every second, and I’m calling that speed. But how much is its speed changing every second? Well if it’s speed was 0 m/s before, and a second later it’s speed is still 0 m/s, then its speed is changing exactly 0 m/s every second, or, it’s *acceleration* is 0 m/s^(2).

So let’s say I’m tired of holding this ball, and I drop it. I want to know what its speed will be 3 seconds later. Well, I know the only thing making it move at all is gravity, and gravity is accelerating it at 10 m/s^2 down (like velocity, acceleration is a vector and has direction, and because gravity pulls down we typically say -9.8 m/s^(2), but again for conceptual understanding I’m leaving the signs out of it), so I would use this equation:

vf = vi + (a × t)

vi = 0 m/s; a = 10 m/s^2; and t = 3 s; so plugging that all in:

vf = 0 m/s + (10 m/s^2 × 3 s) = 0 m/s + 30 m/s = 30 m/s

So, that’s one equation, and it tells me the ball will be moving 30 m/s three seconds after I drop it. But let’s just jump straight to the hardest question: how long will it take for the ball to hit the ground? For this I need this equation:

s = (vi × t) + (0.5 × a × t^(2))

s is displacement, and I know the Empire State Building is 381 m tall, so that is how far the ball has to fall to hit the ground. vi is 0 m/s, and anything times zero is zero, so the first term can go away. a is 10 m/s^2 and we want to know t. So, plugging all of that in we get:

381 m = (0 m/s × t) + (0.5 × 10 m/s^2 × t^(2))

381 m = 5 m/s^2 × t^2

t^2 = 381 m ÷ 5 m/s^2 = 76.2 s^2

t = √(76.2 s^(2)) = 8.7 s

So, it will take the ball 8.7 seconds to hit the ground if I dropped it from the top of the Empire State Building.

Hopefully that helps you understand a little better in terms of kinematics.

Anonymous 0 Comments

Subbing the units and then naming them can help.
Velocity aka speed: V= d/s (change in distance per second)
Acceleration: A=d/s^2= V/s (change in speed per second. / rate of change in speed / how fast your speed is changing for each second that passes)
Does that help conceptualize it?

Anonymous 0 Comments

Subbing the units and then naming them can help.
Velocity aka speed: V= d/s (change in distance per second)
Acceleration: A=d/s^2= V/s (change in speed per second. / rate of change in speed / how fast your speed is changing for each second that passes)
Does that help conceptualize it?

Anonymous 0 Comments

Acceleration measures how fast velocity or speed are changing (velocity is speed in a certain direction, speed doesn’t care about direction).

In your specific example, the car’s acceleration is increasing its speed by 4metres per second every second.

As you noted it starts from rest, this means after 1 second it is moving at 4m/s, after 2 seconds it’s moving at 8m/s and so on up to the 10 second mark where it is moving at 40m/s.

Anonymous 0 Comments

Acceleration measures how fast velocity or speed are changing (velocity is speed in a certain direction, speed doesn’t care about direction).

In your specific example, the car’s acceleration is increasing its speed by 4metres per second every second.

As you noted it starts from rest, this means after 1 second it is moving at 4m/s, after 2 seconds it’s moving at 8m/s and so on up to the 10 second mark where it is moving at 40m/s.

Anonymous 0 Comments

This means the car is quadratically increasing in its *position*. In this case it probably makes the most sense to visualize it as something like a drag race, where time = 0 at the start of the race and position = 0 at the starting line.

Taking a unit “per second” allows you to describe the rate of change of that unit. At a given point in time, we can describe the position of the car in meters past the starting line. If we want to describe the rate at which the position is changing, we can describe a speed in meters per second. That is to say, if the speed is 4 m/s, then the car’s position increases by 4 for every second that elapses. If the car travels at 4 m/s for 10 seconds, then the car’s position will have increased by a total of 40 meters.

Describing movement as “increasing position” is kinda weird but it allows us to make the extrapolation to acceleration much easier.

However, the speed of the car is not constant. We know this because the car is stationary at the start, and moving at the finish line, so the speed changes over the duration. The rate at which the speed changes is what we call acceleration. If we want to describe the rate at which the speed is changing, we can use meters per second per second, which is the same as meters per second squared. That is to say if the car’s acceleration is 4 m/s^2, then the car’s speed increases by 4 m/s for every second that elapses. If the car accelerates at 4 m/s^2 for 10 seconds, then the car’s speed will have increased by 40 m/s.

Now if you want to know how far a car which starts stationary and accelerates at 4 m/s^2 for 10 seconds has gone, then you need to have a bit of an understanding of basic calculus. However, in this case it comes out to be 1/2 * 4 * 10^2 = 200 meters.

Anonymous 0 Comments

This means the car is quadratically increasing in its *position*. In this case it probably makes the most sense to visualize it as something like a drag race, where time = 0 at the start of the race and position = 0 at the starting line.

Taking a unit “per second” allows you to describe the rate of change of that unit. At a given point in time, we can describe the position of the car in meters past the starting line. If we want to describe the rate at which the position is changing, we can describe a speed in meters per second. That is to say, if the speed is 4 m/s, then the car’s position increases by 4 for every second that elapses. If the car travels at 4 m/s for 10 seconds, then the car’s position will have increased by a total of 40 meters.

Describing movement as “increasing position” is kinda weird but it allows us to make the extrapolation to acceleration much easier.

However, the speed of the car is not constant. We know this because the car is stationary at the start, and moving at the finish line, so the speed changes over the duration. The rate at which the speed changes is what we call acceleration. If we want to describe the rate at which the speed is changing, we can use meters per second per second, which is the same as meters per second squared. That is to say if the car’s acceleration is 4 m/s^2, then the car’s speed increases by 4 m/s for every second that elapses. If the car accelerates at 4 m/s^2 for 10 seconds, then the car’s speed will have increased by 40 m/s.

Now if you want to know how far a car which starts stationary and accelerates at 4 m/s^2 for 10 seconds has gone, then you need to have a bit of an understanding of basic calculus. However, in this case it comes out to be 1/2 * 4 * 10^2 = 200 meters.

Anonymous 0 Comments

It’s a measure of change in speed over time.

It means that every second, the speed increases by 4 m/s.

At t=0s, car speed is 0 m/s
at t=1s the speed is 4 m/s
at t=2s speed is 8 m/s
at t=3s speed is 12 m/s

at t=10s speed will be 40 m/s

If the car is not accelerating (ie. speed is constant) then the speed will be the same at every point in time.

Think of it this way, speed is meters per second, ie the change in distance (m – unit of distance) over time (s).
Acceleration is change in speed (m/s – unit of speed) over time (s), so it’s (m/s)/s which then gets shortened to m/s^2

Anonymous 0 Comments

It’s a measure of change in speed over time.

It means that every second, the speed increases by 4 m/s.

At t=0s, car speed is 0 m/s
at t=1s the speed is 4 m/s
at t=2s speed is 8 m/s
at t=3s speed is 12 m/s

at t=10s speed will be 40 m/s

If the car is not accelerating (ie. speed is constant) then the speed will be the same at every point in time.

Think of it this way, speed is meters per second, ie the change in distance (m – unit of distance) over time (s).
Acceleration is change in speed (m/s – unit of speed) over time (s), so it’s (m/s)/s which then gets shortened to m/s^2