# What is the relationship between velocity time and distance

Here's a word equation that expresses the relationship between distance, velocity and time: Velocity equals distance travelled divided by the time it takes to get. Velocity is defined as speed of an object in a particular direction and is expressed as distance moved by it per unit of time, ms−1orms. The relation between velocity and time is a simple one during uniformly A car accelerating for two seconds would cover four times the distance of a car.

So of course if you know two velocities you know more than if you just know one. In the formula for distance: How do you calculate for distance then? You'll have to specify this a little more before we can answer. Is there constant acceleration until that velocity is reached, then the acceleration stops?

If so, I bet you could solve it yourself. Or is there, more plausibly, one of these other situations which also lead to limiting velocities: This applies to objects whose terminal velocities correspond to small Reynold's numbers.

This applies to objects whose terminal velocities correspond to larger Reynold's numbers, including typical large falling objects. Some other effect not in the list? I think you're looking too much into my question.

I don't understand what 'reynold's numbers' are. Or the time be if distance is given, but not time? The last part of this equation at is the change in the velocity from the initial value. Recall that a is the rate of change of velocity and that t is the time after some initial event.

Rate times time is change.

### Distance, Velocity and Time: Equations and Relationship | Science project | catchsomeair.us

Move longer as in longer time. Acceleration compounds this simple situation since velocity is now also directly proportional to time.

Try saying this in words and it sounds ridiculous. Would that it were so simple. This example only works when initial velocity is zero.

Displacement is proportional to the square of time when acceleration is constant and initial velocity is zero. A true general statement would have to take into account any initial velocity and how the velocity was changing. This results in a terribly messy proportionality statement.

• The equation
• Discussion

Displacement is directly proportional to time and proportional to the square of time when acceleration is constant. And this right over here is a change in velocity.

Distance (position) to Velocity Time Graph Physics Help

When we plot velocity or the magnitude of velocity relative to time, the slope of that line is the acceleration. And since we're assuming the acceleration is constant, we have a constant slope. So we have just a line here. We don't have a curve.

Now what I want to do is think about a situation. Let's say that we accelerate it one meter per second squared. And we do it for-- so the change in time is going to be five seconds.

And my question to you is how far have we traveled? Which is a slightly more interesting question than what we've been asking so far. So we start off with an initial velocity of zero. And then for five seconds we accelerate it one meter per second squared. So one, two, three, four, five. So this is where we go. This is where we are. So after five seconds, we know our velocity.

Our velocity is now five meters per second. But how far have we traveled? So we could think about it a little bit visually. We could say, look, we could try to draw rectangles over here. Maybe right over here, we have the velocity of one meter per second.

## Equations of Motion

So if I say one meter per second times the second, that'll give me a little bit of distance. And then the next one I have a little bit more of distance, calculated the same way. I could keep drawing these rectangles here, but then you're like, wait, those rectangles are missing, because I wasn't for the whole second, I wasn't only going one meter per second.

So I actually, I should maybe split up the rectangles. I could split up the rectangles even more. So maybe I go every half second.

So on this half-second I was going at this velocity. And I go that velocity for a half-second. Velocity times the time would give me the displacement.

And I do it for the next half second. Same exact idea here. Gives me the displacement. So on and so forth. But I think what you see as you're getting-- is the more accurate-- the smaller the rectangles, you try to make here, the closer you're going to get to the area under this curve.

And just like the situation here. This area under the curve is going to be the distance traveled. And lucky for us, this is just going to be a triangle, and we know how to figure out the area for triangle. So the area of a triangle is equal to one half times base times height.