How Long Does It Take for Velocity to Change Under Constant Acceleration?

[Tipolymer44]

Homework Statement

I need help on this homework problem I can't 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?

Homework Equations

This is the eq i used: delta T =(Vf-Vi)/a
I set Ti as 0.

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?

 

[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 = { v_{f,along y-axis} - v_{i, 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

 

[Tipolymer44]

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

 

[quietrain]

yes.