# How can I add an acceleration vector to a velocity with a different direction?

I am in my last year of high school and am struggling with some homework. I'm sorry if this question is incredibly stupid, but I simply can't find the answer in my notes.

If I have an object with a velocity of 10ms-1 travelling at a bearing of 090 degrees (to the right) with no acceleration, and then all of a sudden it spontaneously gains an acceleration of 2ms-2 at a bearing of 180 degrees (downwards), how do I calculate the velocity (magnitude and direction) of the object for every second after the object gains this acceleration?

Thank you very much for your help, I really want to be able to understand this!

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You can handle the vertical and horizontal components of the motion seperately, i.e., there is a horizontal velocity and acceleration $v_h, a_h$ and a vertical velocity and acceleration $v_v, a_v$, which seperately obey the usual equations. –  Michael Brown Mar 3 '13 at 8:40
How do velocity and acceleration relate? –  Bernhard Mar 3 '13 at 8:45
Maybe to help you a bit in getting a feel for what this would be in a real life situation: this is equivalent to driving your car off a cliff at 36 km/h with the cliff located on the moon (which has $g=1.62 m/s^2 \approx 2 m/s^2$) –  Michiel Mar 3 '13 at 9:11
@Aeaex This is very much like a question your teacher might have done in the context of horizontal projectile motion and is probably in your notes? It goes something like this: An object is projected horizontally at speed 10ms$^{-1}$ from the top of a tower, find the velocity and its direction of at some time later. The difference between that problem and the one you are showing in this forum, is that the motion is taking place in the x-y plane and you have a=2ms$^{-2}$, instead of g (which is 9.8ms$^{-1}$.) The vector addition for the velocity is shown user John Rennie answer. –  JKL Mar 3 '13 at 10:29

Suppose your object was initially stationary, then using the usual $x-y$ co-ordinates the initial velocity of the object is (0, 0) and the acceleration is (0, -2) i.e. the $x$ component of the acceleration is zero and the $y$ component is 2 m/sec$^2$ downwards. Calculating the velocity as a function of time is straightforward because the object accelerates down the $-y$ axis so the velocity at some time $y$ is simply:

$$\vec{v}(t) = (0, -2t)$$

The only difference between the stationary object and the one in your question is that in the question the object starts out moving at 10 m/sec to the right, i.e. $\vec{v}$(0) = (10, 0)

so to calculate the velocity of the object in your question you just have to do a vector sum of the initial velocity and the change in the velocity due to the downward acceleration.

Note that the acceleration changes both the magnitude and the direction of the initial velocity. This is quite general, and for example it explains why satellites orbit the Earth. At any instant of time their velocity is at right angles to the line joining them to the Earth and the acceleration due to gravity acts along the line. This is just like your question.

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You will calculate the velocity for the x, y and z components separately.

So if the object has $v_x$ velocity initially, and it spontaneously gains acceleration in the negative y direction, this problem becomes very simple. Over time this object will gain velocity in the y ($v_y$) direction. It will also always have its' velocity in the x direction, and that velocity will never change, what I mean is that $v_x$ is a constant.

In university you will learn to separate a vector into its different components, which helps simplify things a lot. So I will use this notation to explain, hopefully you understand.

$v_0 = (10m/s) \hat{i}$

$a_0 = -(2m/s^2) \hat{j}$

$v_t = (10m/s)\hat{i} - (2m/s^2\cdot t)\hat{j}$

So you would simply plug in the value of time and you would have the individual components of the velocity at any time, including at t=0.

To find the magnitude of velocity you would use the pythagorean theorem.

$r = \sqrt{x^2+y^2}$

and you can use the arctangent function to find the angle

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