When/where will two projectiles converge

1. Jul 23, 2010

Fortson1

1. The problem statement, all variables and given/known data

Two projectiles, projected at the same time with the following attributes:

Object 1: Traveling 7 feet/per sec, accelerating -.25 feet/per sec
Object 2: Starting projected position is 80 feet ahead of object 1, traveling 4 feet/per sec, accelerating -.45 feet/per sec.

How far will these objects travel before they meet and how much time will it take for them to meet.

(As these objects slow, they can even reverse direction.)

2. Relevant equations

$$d=v*t$$

3. The attempt at a solution

$$d=7fps*t$$

$$d+80=4fps*t$$

I'm not a physics student and unfortunately never was near that level of math, and my ability to take a further stab at this is rather limited. This is a generalized problem that I'm trying to learn how to resolve from my workplace.

I am very lost in how to set this up but really want to learn!

Thanks guys (and gals) for any assistance.

Last edited: Jul 24, 2010
2. Jul 24, 2010

kuruman

The relevant equation for each projectile under constant acceleration is

x(t)=x0+v0t+(1/2)at2

The meaning of the symbols is
x(t) = position of the object at any time t.
x0 = position of the object at the specific time t = 0.
v0 = velocity of the object at time t = 0.
a = acceleration of the object.
t = any time t.

Write one such equation for each of the objects in the form
x1(t) = ....
x2(t) = ....
These equations give you the position of each object at any time t.
Now there exists at specific time t = tm at which the objects meet. How would you express the event that they meet at that time with an equation?

3. Jul 24, 2010

Fortson1

Thanks so much kuruman for getting me started on this. Here goes:

Object 1:
x0 = 0 (if time is "0", it hasn't gone anywhere thus position would be "0")
v0 = 7f/s
a = -.25f/s
t = unknown

Object 2:
x0 = 80
v0 = 4f/s
a = -.45f/s
t = unknown

Therefore (if this is correct) the equations for each object would be:

object 1:) x(t)=0 + 7f/st + ½*-.25t2
object 1:) x(t)=80 + 4f/st + ½*-.45t2

For example, if t= 12 (seconds)

object 1:
x(12)=0+7f/s(12) + ½*-.25(122)
= 84 + (-18)f
= 66 feet after 12 seconds

object 2:
x(12)=80+4f/s(12) + ½*-.45(122)
=80+48 + (-32)
=96 feet after 12 seconds

In order to express at what time these two objects will meet, tm, we know that t for both equations will be equal when this occurs. So I'm thinking I would need to solve for "t" for each equation, and then make the results equal to each other? (if so, I'm not quite sure how to move everything over when there is "t" and "t2", etc.)

4. Jul 24, 2010

kuruman

Good job!

Now when the two objects meet, this means that they are in the same position at the same time. So if you replace any time t in each of the equations with tm, and you calculate the position, you should get the same number! So you need to say that

x1(tm) = x2(tm)

and solve the resulting equation for tm. Can you do that?

5. Jul 24, 2010

Fortson1

Awesome, thanks.

The initial reference that you made to 'any time t':

x(t)=x0+v0t+(1/2)at2

so replacing (t) with (tm) would be:

X1(tm) = x0+v0t+(1/2)at2
X2(tm) = x0+v0t+(1/2)at2

and as you noted,

X1(tm)=X2(tm), then

x0+v0t+(1/2)at2 = x0+v0t+(1/2)at2 (I'm not sure how to properly express this to differentiate object 1 variables from object 2 variables)

If I plug in the numbers, I get

7t + ½*-.25t2 = 80 + 4t + ½*-.45t2
7t + (-.125)t2 = 80 + 4t - .225t2
7t + (-.125)t2 - 80 - 4t + .225t2 = 0
7t - 4t -.125t2 + .225t2 - 80 = 0
3t + .1t2 - 80 = 0
+.1t2 + 3t - 80 = 0

a = .1
b = 3
c = -80

(sorry for the lack latex knowledge...)

For the quadratic equation, I get (-3 +/- sqrt(9+32)), divided by .2

yielding 17, -47
But since time (in this case) time can't be negative,

I've plugged it in and it adds up. This took me a very long time to resolve...not because I was so confused, but because I kept making +/- and decimal mistakes when calculating and moving the numbers, etc. I'd love to call that math-rust, but I was plagued by it decades ago in school too. :tongue:

Thanks again, kuruman! If you have anything to add, of course please do. This has been a fantastic learning experience. I will also work on using latex to make any future questions and replies a bit easier to follow.

/edit: Amidst my enthusiasm for finally solving it, I completely overlooked the 2nd part of my problem, which is "where" will they meet. But while verifying the answer of "17" by plugging it into each equation, I inadvertently took care of this and found that the answer is 83 feet (as each equation properly resulted in the same answer.)

Last edited: Jul 24, 2010
6. Jul 25, 2010

kuruman

Excellent work, Fortson1!
There is one minor detail that I should point out about your comment
When you write
x1(tm)=x2)tm),
the next line (symbolically) should be
x01+v01tm+½a1tm2=x02+v02tm+½a2tm2.
Note the use of subscripts to differentiate the symbols associated with each object. This helps to sort things out when you put in numbers. Also note that "any time t" has been replaced with "specific meeting time tm". This use helps sort out the times (if you have a more complicated problem) at which things happen. I like to think of "any time time t" as an independent variable, a placeholder, until it is appropriate to replace it with something related to the problem.

Of course, these are minor points intended for future reference and do not in any way imply that your solution is incorrect. Furthermore, I applaud your checking your solutions by substituting your numerical answers back into the equations to verify consistency. Quite a few people accept the first number that comes out of their calculator without this simple check.

7. Jul 25, 2010

Fortson1

Thanks again.

I very much appreciate all of your tips (and can't emphasize that enough.) I was actually hoping to get some feedback referencing variables without wanting to overly-bombard you for assistance.

However there is something else I'd like to check if you don't mind.

A thought came to mind that might be useful for me in the future is calculating when an object with a constant acceleration (such as the properties of these objects) 'reverses direction', or reaches the point where it would reverse direction.

I got some insight into this when plugging previous results into the equations to see if they are correct and am thinking the relevant equation would be:

v0t=(1/2)at2

If this is correct, then I would solve for t:

v0t=(1/2)at2
(v0t)/a=(1/2)t2....mult. each side by 1/a
(2v0t)/a=t2.........mult. each side by 2
2v0/a=t..............................................after mult. each side by 1/t

But if I plug my Object 1 numbers in (v0=7, a=-.25), I get:

(2*7)/-.25=t
-56=t

My concern here is that 't' is negative.

Is this the correct approach or am I missing something?

Last edited: Jul 25, 2010
8. Jul 25, 2010

kuruman

I don't mind.
[/QUOTE]
Good question. You need to understand the meaning of "velocity" which is not the same as "speed" in physics. Velocity has direction, but speed does not. So if we adopt the convention that "up" is positive and "down" is negative, an object that is moving up has positive velocity (and positive speed) and an object that is moving down has negative velocity (and still positive speed). Now imagine throwing a ball straight up in the air. It moves under constant acceleration which is a = -32 ft/s2. The minus sign means that the acceleration is in the down direction (as we know it is.) The ball's velocity at any time t is given by

v(t) = v0 +at = v0 +(-32 ft/s2)t

The equation is saying that "for every one-second interval, you have to subtract 32 ft/s from the velocity the ball had at the beginning of the interval."

Now let's put it together logically
1. As long as the ball is moving up, the velocity is positive.
2. As long as the ball is moving down, the velocity is negative.
Therefore when the ball reverses direction the velocity must be _______.

A ball is shot straight up with initial speed 48 ft/s. At what time, after launch, does the ball reverse direction?

9. Jul 25, 2010

Fortson1

I see, that's good imagery and easy to follow. When a ball reverses, it will have 0 velocity (and 0 speed, if it was thrown straight up, otherwise it would be a function of trajectory). So if I’m not mistaken, I’m looking to solve for time when velocity=0.

In the case of your example:

v0=48ft/s
a =-32ft/s2

therefore,

48ft/s+(-32ft/s2)t=0
-32ft/s2t=-48ft/s
t=(-48ft/s)/(-32ft/s2)
t=3/2, or 1.5 seconds

And to solve my problem for Object 1:

v0=7ft/s
a =-.25ft/s

7ft/s + (-.25ft/s)t = 0
-.25t=-7
t=-7/-.25
t=28 feet!

One question: you referenced 32ft/s2...and that is different from 32ft/s without the seconds being squared? And does this seconds-squaring commonly get factored or simplified, etc. as well...for example it is divided by 5ft/s, without a square (or is that situation unusual)?

10. Jul 26, 2010

kuruman

Velocity is the rate of change of position with respect to time. Its units are ft/s, so that if you say that the velocity is 32 ft/s you mean that "for every second that goes by, the position changes by 32 feet."
Acceleration is the rate of change of velocity with respect to time. Its units are ft/s/s, so that if you say that the acceleration is 32 ft/s/s you mean that "for every second that goes by, the velocity changes by 32 feet per second."

Conventionally, we write ft/s/s as ft/s2. More generally, the dimensions of velocity are

$$[v]=\frac{Length}{Time}$$

and of acceleration
$$[a]=\frac{Length/Time}{Time}$$

where "Length" would be your choice of feet, meters, inches, light-years, etc. and "Time" would be seconds, hours, years, etc.

11. Jul 26, 2010

Fortson1

That's interesting and very straight forward spelled out, thanks, kuruman. That's one less thing for me to get tripped up on at some point in the future, which is always a good thing.