Q:

Suppose a car travels at a constant speed of 10m/s for 4 seconds. Then his total distant will be 40 meters. Now if the car travels at a uniform accelaration of 2.5m/second squared for 4 seconds and his final velocity is 10m/s, then his total distance will be 20 meters. WHY?

- rezwan (age 13)

dhaka, bd

- rezwan (age 13)

dhaka, bd

A:

Rezwan -

In your first example, you were able to calculate the car's final position by multiplying its velocity (speed) by the time in which it moved. This is because velocity is how much the position changes in a certain amount of time, or change in position divided by change in time. If we say that the car starts at position zero and time zero, then velocity = final position / time. Another way of writing this is velocity * time = final position. Or 10 m/s * 4 seconds = 40 meters.

The same happens when you start with acceleration, except that you have to do it in 2 steps. In the first step, you have the acceleration, which is the change in velocity over a certain amount of time. Again, we can write this by saying that acceleration = final velocity / time. Another way of writing this is to say that final velocity = acceleration * time. Or as in your example, 10 m/s = 2.5 m/s^2 * 4 seconds.

Now it seems like we should be able to calculate the distance from the velocity just like before. Or can we? In the first case, the velocity wasn't changing. So we could just multiply it by time. But to find the final position (the distance traveled), we have to use the*average* velocity of the car. If the velocity doesn't change, then the average velocity is the same as the final velocity. But if the velocity is increasing with time, then the average velocity will be less than the final velocity.

In this case, because the acceleration is constant, it turns out that the average velocity is half of the final velocity, or 5 m/s. If this doesn't make sense to you, try drawing a graph of the velocity versus time. Initially, the velocity is zero. And after 4 seconds, it's 10 m/s. So the average velocity is the velocity right in the middle, at 2 seconds, or 5 m/s.

Now as the second step, we can use this velocity to calculate the final position: Average velocity * time = final position, or 5 m/s * 4 seconds = 20 meters.

-Tamara

In your first example, you were able to calculate the car's final position by multiplying its velocity (speed) by the time in which it moved. This is because velocity is how much the position changes in a certain amount of time, or change in position divided by change in time. If we say that the car starts at position zero and time zero, then velocity = final position / time. Another way of writing this is velocity * time = final position. Or 10 m/s * 4 seconds = 40 meters.

The same happens when you start with acceleration, except that you have to do it in 2 steps. In the first step, you have the acceleration, which is the change in velocity over a certain amount of time. Again, we can write this by saying that acceleration = final velocity / time. Another way of writing this is to say that final velocity = acceleration * time. Or as in your example, 10 m/s = 2.5 m/s^2 * 4 seconds.

Now it seems like we should be able to calculate the distance from the velocity just like before. Or can we? In the first case, the velocity wasn't changing. So we could just multiply it by time. But to find the final position (the distance traveled), we have to use the

In this case, because the acceleration is constant, it turns out that the average velocity is half of the final velocity, or 5 m/s. If this doesn't make sense to you, try drawing a graph of the velocity versus time. Initially, the velocity is zero. And after 4 seconds, it's 10 m/s. So the average velocity is the velocity right in the middle, at 2 seconds, or 5 m/s.

Now as the second step, we can use this velocity to calculate the final position: Average velocity * time = final position, or 5 m/s * 4 seconds = 20 meters.

-Tamara

*(published on 10/22/2007)*

Q:

Is there a mistake in the units in the answer? I believe in figuring the average velocity from the acceleration, the velocity should be in m/s and the acceleration in m/s ^2.

- Levin Tull (age 56)

San Antonio, TX, USA

- Levin Tull (age 56)

San Antonio, TX, USA

A:

Yipes- No there's no mistake there now. That's because, in response to your note, I fixed it.

It's hard to believe that sat around on this site for years and nobody noticed.

Many thanks!

Mike W.

It's hard to believe that sat around on this site for years and nobody noticed.

Many thanks!

Mike W.

*(published on 02/28/2011)*