# 1-D Motion

1. Jun 8, 2005

### Naeem

Q. An object starts at time t = 0 with a velocity of v0 = + 21 m/s and undergoes a constant acceleration of a = -5 m/s^2.

a) How far from its starting ( t = 0 ) is the object at time at which it comes to rest?

Solution:

Since, the object undergoes constant acc. constant acceleration equations may be applied.

Therefore,

We need to find the time 't' first,

so, v = v0 + at ( It is going to rest , so v = 0 m/s )

-v0 = at

Given a = -5 m/s^2.

Plugging in we get t = 4.2 seconds, which I believe is right.

Then, using:

x -x0 = v0t - 1/2 a*(t2-t1)^2

x0 ----> 0
v0-----> +21 m/s
't' from above ------> 4.2 seconds which is 't2) and t1 = 0 ---> at rest.
'a' from above -------> -5 m/s^2

Plugging in all the know data in the above eqn, then give us.

x = (21 m/s)*(4.2 seconds) - 1/2 ( -5 m/s^2) * ( t2-t1)^2

I got x = 88.2 + 44.1 = 132.2 meters.

Is this correct!

b) Suppose a second object begins moving with a constant speed of v = 19 m/s in the same direction from the same location 3 seconds later. At what time do the paths of these two objects again cross?

If the objects cross, then their x - coordinates, may be set equal to one another and solved for time. That's all I know. I tried to setup by doing the following:

1st case:

v0 = +21 m/s
a = -5 m/s^2
t = 0 at rest

2nd case :

v0 = 19 m/s
and since its constant speed acc. - 0.

setting up the constant acc,equations for the 1st case data:

x = (21m/s)t - 1/2 (-5 m/s^2)t^2

setting up the constant acc, equations for the 2nd case data:

x = (19m/s)t [ and the rest of the terms vanish since speed is constant for the second case )

equating the above we get:

21t + 2.5t^2 = 19t

2.5t^2-19t+21t = 0

is this correct so far. Is my reasoning correct for both the parts so far.
and my work!

Can one be kind enough to tell me if something is wrong.

Thanks, for looking
and appreciate some feedback!

2. Jun 8, 2005

### whozum

A)

If you use a = -5, then you must place a positive instead of the bolded negative above. If you use a = 5, then you can keep it the way it is.

B)

Same thing as above.

Your reasoning is correct for this portion, but you need to take into account that the first puck has a 3 second lead! Also note, that they will cross when the first puck is on its way BACK.

3. Jun 8, 2005

### learningphysics

The above "presumes" that acceleration is in the negative direction. Like whozum said if you use a=5 then the above is ok. I'd advise you to just use

in general... and then plug in the appropriate a. Here you'd plug in a=-5.

Also for your first part... there is a different equation you can use to solve immediately.

d=(v2^2 - v1^2)/(2a)

4. Jun 9, 2005

### Naeem

For part b, that means ,change it as follows:

x = (21m/s)t - 1/2 (-5 m/s^2)(t-3)^2

or change the 2nd case, equation, to x = (19m/s)(t-3)

5. Jun 9, 2005

### OlderDan

Too many minus signs in your first equation. The general formula is

x = x_0 + (v_0)t + (1/2)at^2

Also, the first object starts from x = 0 at t = 0. There should not be a (t-3) in the first equation. The second equation (for the second object) is correct. The two objects will meet when x is the same in both equations. Set your two expressions for x equal to one another and solve for t.

6. Jun 9, 2005

### whozum

Do you have a TI-83? Put it in parametric mode and graph the following to visualize the problem:
X1 = T
Y1 = -5T^2/2 + 19T
X2 = T
Y2 = 21(T-3)