# Of Velocity, Acceleration, and Movement

1. Jun 23, 2005

### senkyoshi

NOTE: I have a physics question that I came across while working on a personal project. This is not a homework question. I was not sure which forum to put his in. I am not great at physics. However, I find that once in a while I need to use it in my daily life. I should have listened to my physics teacher when they told us that in class!

Say I have an object that will move 5 inches in one direction in a straight line. It will also move with a velocity parameter of 7 inches/sec and an acceleration parameter of 10 times that of the velocity (70). What I need to know is how many inches will it take for the object to reach velocity? And knowing that, how many seonds will that be?

This is my first post in this forum so I do not know exactly how things work here but I do not want only an answer, I want to understand how this works as well so I can do it in the future and/or apply it to other projects ("Teach a man to fish...").

2. Jun 23, 2005

### Staff: Mentor

Does the object undergo constant acceleration with an initial speed of 0? Is your question: Starting from rest, given the constant acceleration, how much distance and time does it take for the object to reach the given speed?

3. Jun 23, 2005

### senkyoshi

Doc Al,

Yes, I think you get what I was trying to say. Thanks.

4. Jun 23, 2005

### HallsofIvy

The speed of any object, with contant acceleration a, initial velocity v0, will have velocity v(t)= at+ v0 after t seconds. During that time it will have moved a distance d(t)= (1/2)at2+ v0t.

In your case, since the initial velocity is v0= 0 and the constant acceleration is a= 70 in/sec, the velocity after t seconds will be v(t)= 70t and the distance moved will be d(t)= 35t2. To find the time necessary for the object to accelerate from 0 to 7 inches per second, solve the equation 70t= 7. To find the distance moved in that time substitute that answer for t in d(t)= 35t2.

(It won't take very long and it won't go very far!)

5. Jun 23, 2005

### senkyoshi

HallsofIvy,

Thanks for your response. Here is what I get. Check me if I am wrong.

For Time:
70t = 7
t = 7/70 = 0.1
For Distance in inches:
d(t) = 35t^2
= 35(0.1)^2
= 35(0.01)
= .35

I think I got it. Can you explain where you got the first 2 equations that are in your response for Velocity and Distance?
Thanks a bunch!

6. Jun 23, 2005

### Staff: Mentor

The first equation that Halls' provided is essentially a statement of what acceleration means, the rate of change of velocity. Since the acceleration is uniform:
$$a = \frac{\Delta v}{\Delta t}$$
This can be rewritten as:
$$v_f = v_0 + a t$$

The second equation combines that first equation with the definition of average speed as distance divided by time. The average speed during the motion is $v_{ave} = (v_0 + v_f)/2 = (v_0 + v_0 + a t)/2$. Thus the distance will equal the average speed X time:
$$x = (v_0 + v_0 + a t)/2 \times t = v_0 t + 1/2 a t^2$$

Make sense?

7. Jun 23, 2005

### senkyoshi

I think I've got it. Thanks guys! I did a little research of my own and found the first equation but the one about Distance threw me off.

8. Jun 23, 2005

### senkyoshi

Okay, maybe I dont understand all the way. Can one of you please derive the distance equation(d(t)= (1/2)at2+ v0t) for me step by step? I would really appreciate it. I cannot make it all the way. Thanks.

9. Jun 23, 2005

### Staff: Mentor

I'll break it down in steps. Tell me where I lose you.
(A) Find the average speed in terms of initial and final speeds: $v_{ave} = (v_0 + v_f)/2$

(B) Plug equation 1 ($v_f = v_0 + at$) into the average speed equation just found: $v_{ave} = (v_0 + v_f)/2 = (v_0 + v_0 + a t)/2$. Simplify to: $v_{ave} = (2v_0 + a t)/2 = v_0 +1/2 a t$.

(C) Use distance = average speed X time. Plug in what we just found for average speed and multiply by time:
$x = v_{ave} \times t = (v_0 +1/2 a t) \times t = v_0 t + 1/2 a t^2$.

Last edited: Jun 23, 2005
10. Jun 23, 2005

### senkyoshi

OH! When I was reading it I missed the $v_0 + v_0 = 2v_0$ part.

Makes sense now. thanks again!