# Homework Help: 2-Dimensional Motion Problems

1. Oct 14, 2004

### Phyzix

OK---a rocket is launched at 53 degrees above horizontal with an initial speed of 100 m/s. It moves along its initial straight line path with an acceleraqtion of 30 m/s^2 for 3 seconds. At this time, the rocket is turned off and it proceeds to fall as a free body.

how can i find,A) the maximum height reached. B) the total time of flight. C) its horizontal range. ???????

another problem im having is the escalator problem which the vectors move in the same direction...

You notice you can walk up a stationary (broken) escalator in 30 seconds-- while you can go up in 20 seconds while it is working just by standing on it. How long would it take for you to WALK up the working escalator?

any help plz?!

2. Oct 14, 2004

### aekanshchumber

Second problem,
Let the total distance be S, your speed V1 and speed of esculator is V2
30 = s/V1 and, 20 = s/V2
when you are walking on working esculator let t be the time taken,
t = s/(V1 + V2)
or, 1/t = 1/30 + 1/20 = 1/12
t= 12sec

(also tell that in first question, acceleration is initial or at the point where engine is turned off.)

Last edited: Oct 14, 2004
3. Oct 14, 2004

### Phyzix

the acceleration in problem 1 is the acceleration during the period of time when the rocket is on--not when it has been turned off...

btw, thanks a lot for the help with the other one but

t=S/(V1+V2)....where did you get 1 for S....did you just use it as 1 unit? and why does the t become 1/t?

4. Oct 14, 2004

### aekanshchumber

No, S is not replaced by 1, here is detail.
t = s/(V1 + V2)
taking inverse
1/t = (V1 + V2)/s = V1/s + V2/s = 1/30 + 1/20.

First one.
I am solving it for a genral solution, v be the initial velocity, h be the max. height, a be the acceleration, t be the time during which it accelerated, x is angle of projection.
h1 = vt - 0.5(g-asinx)t^2
(g - asinx) net acceleration in downward direction.
h2 = [(vf^2)/2g] where vf is the final velocity when power is off.
h = h1 + h2
Total time of flight T,
This one is a shortcut tech. max. height attained h, let the body is droped from heigh h and then solve it for time taken
t1 = (2h/a)^0.5 a is total acceleration.
t1 is the time taken to fall, but under standerd condition time taken to attain the height h willalso be t1
T = 2*t1

Horizontal range R,
R = (horizontal velocity)*total time

Last edited: Oct 14, 2004
5. Oct 14, 2004

### Phyzix

Thanks, I think that I understood that really well....but a similar question is:

A spaceship is observed traveling in positive x direction with a speed of 150 m/s when it begins accelerating at a constant rate. The spaceship is observed 25 s later traveling with an instantaneous velocity of 1500 m/s at an angle of 55 degrees above the x axis. What is the magnitud of acceleration of the spaceship during the 25 seconds?

If i make the triangle i have 1500-150/t=a. This gives me 1350/25=a which is 54. However, i've been told that is wrong...I'm pretty sure that it has to do with the change in direction of the spaceship but i'm lost as in where to go...

thanks again!

6. Oct 15, 2004

### aekanshchumber

In this problem the change in velocity is in two dimension (X and Y). To calculate the acceleration we need change in velocity and time taken to change the velocity, only thing to do is to have correct equations,
Calculate horizontal acceleration using horizontal velocity and vertical with the vertical velocity, again a genral solution.
a = (V - U)/t V and U are final and initial velosities resp.
taking only horizontal acceleration in account,
acosx = (Vcosx - Ucosx)/t .............1 [x is the angle]
now same with the verticle one,
asinx = (Vsinx - Usinx)/t .............2

using 1 and 2 a can be found easily.
{while trying it, you were on the right track but with the wrong approch, basic concept was right but you couldn't apply it correctly.}