# Another One Dimensional Motion Problem

1. Aug 22, 2009

### pointintime

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

A rock is thrown vertically upward with a speed of 12.0 s^-1 m. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 20.0 s^-1 m.

(a) At what time will they strike each other?

(b) At what height will the collision occur?

(c) Answer (a) and (b) assuming that the rder is reversed: the ball is trown 1.00 s before thr ock.

2. Relevant equations

X = Xo + Vot + 2^-1 a t^2

3. The attempt at a solution
I have no idea how to even start this problem
were the 1 subscripts are the first object that went off and the twos are the second object
t1 = t2 - 1.00s

plugged in

Xo + Vo1 t1 + 2^-1 a t1^2

Xo + Vo1 (t2 - 1.00 s) + 2^-1 a (t2 - 1.00)^2

solved for t2 found 1 second...

2. Aug 22, 2009

### pointintime

if i consider there velocities to be positive do i have to make there accelerations negative???

3. Aug 22, 2009

### pointintime

do they collide at 2.152 s + 1.00 s ?????

not sure if that's right...

4. Aug 22, 2009

### tiny-tim

Hi pointintime!

(try using the X2 and X2 tags just above the Reply box )
If you consider distance upward to be positive, then yes, you have to make the accelerations negative.

(and if you want us to check your result, do show the whole calculation! )

5. Aug 22, 2009

### pointintime

how do you recomend i solve it

6. Aug 22, 2009

### tiny-tim

The method you're using looks fine

7. Aug 22, 2009

### pointintime

X1 = X2
sense they collide

t1 = t2
point at which the collide

t1 = t2 - 1.00 s

X1 = Xo + Vo t + 2^-1 a t^2

X1 = (12.0 s^-1 m)t1 + 2^-1 (9.80 s^-2 m)t1^2

sense
t1 = t2 - 1.00 s

X1 = (12.0 s^-1 m)(t2 - 1.00 s) + 2^-1 (9.80 s^-2 m)(t2 - 1.00 s)^2
X1 = (12. 0 s^-1 m)t2 - 12.0 m + 2^-1 (9.80 s^-2 m)(t2^2 + 1.00 s^2 - 2 t2 1.00s)
X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)(t2^2 + 1.00 s^2 - t2 2.00s)
X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)t2^2 + 4.90 m - (9.80 s^-1 m)t2
X1 = (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m

sense X1 = X2

(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = Xo + Vo t2 + 2^-1 a t2^2
(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = (20.0 s^-1 m) t2 + 2^-1 (9.80 s^-2 m) t2^2
(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m - (20.0 s^-1 m) t2 -(4.90 s^-2 m) t2^2
(17.8 s^-1 m)t2 - 7.1 m

what now???
let me guess I did it wrong

(17.8 s^-1 m)t2 - 7.1 m = 0
(17.8 s^-1 m)t2 = 7.1 m
t2 = .3989 s

what does that mean????

8. Aug 22, 2009

### pointintime

can someone check my work i probably did it wrong and if i didn't i don't know what to do from there

9. Aug 22, 2009

### tiny-tim

Well, it's bit difficult to read (please use the X2 tag in future), but the method looks ok,

except (i thought you agreed on this) the acceleration needs to be negative.

10. Aug 22, 2009

### pointintime

It does have to be negative???
If I consider it to be positive does it miss anything up

11. Aug 22, 2009

### tiny-tim

Yeees! …

if your 12 and 20 are positive, then your 9.8 must be negative (or vice versa).

what's up for velocity must be up for acceleration!​