# Constant accleration problem

1. Sep 13, 2011

### Tipolymer44

1. The problem statement, all variables and given/known data
I need help on this hw problem I cant seem to figure it out.
A objects constant acceleration is north at 40 m/s^2. At time 0 its velocity vector is 15 m/s west at what time will the magnitude of the velocity be 30 m/s?

2. Relevant equations

This is the eq i used: delta T =(Vf-Vi)/a
I set Ti as 0.
3. The attempt at a solution
So I got 30-15/40= .375 secs

I also tried subtracting gravity from acceleration to give (30-15)/(40-9.8) =.496 secs

These are apparently wrong. what should i be doing?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 14, 2011

### quietrain

there is direction involved here so you cannot just pluck values into a formula

so the question nicely sets north (y-axis) and west (x-axis) for you.

so question wants the time when the MAGNITUDE of velocity is 30m/s, which means your x and y components of the velocity, will have to give you a resultant of 30m/s

notice your x-component velocity of 15m/s doesn't change since there is no acceleration in this direction, acceleration is only towards north (y-axis)

so in order to find the y-component velocity THAT will give you 30m/s resultant velocity, you have to use Pythagorean theorem as it is a triangle. draw it out and you will see

so its 302 = 152 + y2, where y is your velocity in y-axis direction

so solving, y = 26m/s

so now you use your formula ALONG THE Y-direction

time taken = { vf,along y-axis - vi, along y-axis } / a along y-axis
which is

(26 - 0) / 40 = 0.65s , the initial velocity is 0 because the object is moving west, there is no y-direction velocity

3. Sep 14, 2011

### Tipolymer44

ahhh, so i basically need to formulate the directional components before i can solve for t

4. Sep 15, 2011

yes.