Acceleration, Velocity, Distance, Time
Most recent answer: 10/22/2007
Q:
How do you find acceleration when given velocity and distance?
- Anonymous (age 17)
Malaysia
- Anonymous (age 17)
Malaysia
A:
The basic equation for solving this is:
d = vt + (1/2)at2
where d is distance traveled in a certain amount of time (t), v is starting velocity, a is acceleration (must be constant), and t is time. This gives you the distance traveled during a certain amount of time. If you know any 3 of those things, you can plug them in to solve for the 4th. So if you only know v and d, you can't solve for a unless you also know what t is (i.e. what time was d measured at).
-Tamara
d = vt + (1/2)at2
where d is distance traveled in a certain amount of time (t), v is starting velocity, a is acceleration (must be constant), and t is time. This gives you the distance traveled during a certain amount of time. If you know any 3 of those things, you can plug them in to solve for the 4th. So if you only know v and d, you can't solve for a unless you also know what t is (i.e. what time was d measured at).
-Tamara
(published on 10/22/2007)
Follow-Up #1: kinematics
Q:
The answer given to this question is incorrect. When you do not know the time, but have the velocity and distance, AND you know it is undergoing constant accelleration you can use the formula
d = 1/2 (Vi+Vf) x t
to solve for t and then you have the 3 required variables to solve the equation given.
to make this a question, am I right?
- Travis (age 24)
Paisley, ON, Canada
- Travis (age 24)
Paisley, ON, Canada
A:
The original answer apparently assumed that the velocity you knew was only the initial one. In that case that answer is correct as stands. You seem to assume we know both the initial and final velocities. In that case we know the average velocity (if the acceleration is constant) vave = (vF+vI)/2 and can then solve for t=d/vave. We then can use a=(vF-vI)/t. So of course if you know two velocities you know more than if you just know one.
Mike W.
Mike W.
(published on 01/23/2008)
Follow-Up #2: acceleration, limiting velocity, distance
Q:
In the formula for distance: d = vt + (1/2)at^2, what if there's a maximum velocity? How do you calculate for distance then?
- Joe (age 21)
- Joe (age 21)
A:
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:
1. An acceleration (e.g. g) reduced by a friction force that's linear in velocity? (This applies to objects whose terminal velocities correspond to small Reynold's numbers.)
2. An acceleration (e.g. g) reduced by a friction force that's quadratic in velocity? (This applies to objects whose terminal velocities correspond to larger Reynold's numbers, including typical large falling objects.)
3. A constant force applied to an object which reaches relativistic speeds, so that F=ma no longer is accurate?
4. Some other effect not in the list?
Mike W.
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:
1. An acceleration (e.g. g) reduced by a friction force that's linear in velocity? (This applies to objects whose terminal velocities correspond to small Reynold's numbers.)
2. An acceleration (e.g. g) reduced by a friction force that's quadratic in velocity? (This applies to objects whose terminal velocities correspond to larger Reynold's numbers, including typical large falling objects.)
3. A constant force applied to an object which reaches relativistic speeds, so that F=ma no longer is accurate?
4. Some other effect not in the list?
Mike W.
(published on 04/01/2011)
Follow-Up #3: maximum velocity?
Q:
I think you're looking too much into my question. I don't understand what 'reynold's numbers' are. :)
V = initial velocity
D = distance over time
T = Time
A = acceleration, a constant
So really, if there's a maximum velocity - what will the distance be? Or the time be if distance is given, but not time? I'm also wondering if the formula gets adjusted at all to compensate for a velocity limit?
- Joe (age 21)
- Joe (age 21)
A:
Here's the thing: If the acceleration remains constant, you can't have a maximum velocity. The velocity will just get bigger and bigger in the direction of the acceleration. So there must be some rule about how the acceleration stops or tapers off to give that maximum velocity.
Forget our remarks about Reynold's numbers etc, just let us know what's supposed to be going on physically. Is this about a ball dropped from a tower? A car driven by a law-abiding motorist? A particle in a linear accelerator? Without that information there's no way we can give an answer.
Mike W.
Forget our remarks about Reynold's numbers etc, just let us know what's supposed to be going on physically. Is this about a ball dropped from a tower? A car driven by a law-abiding motorist? A particle in a linear accelerator? Without that information there's no way we can give an answer.
Mike W.
(published on 04/01/2011)
Follow-Up #4: gravity with friction: distance vs. time
Q:
Okay, how about gravity with terminal velocity, concerning distance and time?
I'm just looking for a formula, and I don't need numbers. If it helps you explain, that's fine too.
- Joe (age 21)
- Joe (age 21)
A:
Ok, I'll do the case of some very fluffy thing where the friction is proportional to the velocity, falling in gravitational acceleration from an initial velocity of zero, reaching terminal velocity VF. Then after time t the height fallen, h, is given by:
h= VFt+(VF^2/g)(e^(-gt/VF)-1).
The more common case, like for a ball, where the friction becomes proportional to the square of the velocity after a short period of acceleration is harder to solve. I might have to use a program, or look it up somewhere, if you want that formula too.
Mike W.
So Mathematica didn't download properly, and I had a chance to sleep on it. For the case where the friction is proportional to velocity squared, here's what I get:
time= (VF/g)*(integral over [1, e^(hg/VF^2)] of dx/sqrt(x^2-1)
With some help from an integral table:
time=(h/VF)+(VF/g)*ln(1+sqrt(1-e^(-2hg/VF^2))).
I checked the limiting cases (small h and big h) and it looks ok.
Addendum- I'd left out the expression for distance h in terms of time, t, thinking it would be even messier. It actually comes out simple:
h=(VF^2/g)*ln(cosh(tg/VF))
Again, limiting cases check.
h= VFt+(VF^2/g)(e^(-gt/VF)-1).
The more common case, like for a ball, where the friction becomes proportional to the square of the velocity after a short period of acceleration is harder to solve. I might have to use a program, or look it up somewhere, if you want that formula too.
Mike W.
So Mathematica didn't download properly, and I had a chance to sleep on it. For the case where the friction is proportional to velocity squared, here's what I get:
time= (VF/g)*(integral over [1, e^(hg/VF^2)] of dx/sqrt(x^2-1)
With some help from an integral table:
time=(h/VF)+(VF/g)*ln(1+sqrt(1-e^(-2hg/VF^2))).
I checked the limiting cases (small h and big h) and it looks ok.
Addendum- I'd left out the expression for distance h in terms of time, t, thinking it would be even messier. It actually comes out simple:
h=(VF^2/g)*ln(cosh(tg/VF))
Again, limiting cases check.
(published on 04/01/2011)