Interpreting the vector addition of acceleration

You can give the answer your own way by taking a situation like: A particle originally at rest has acceleration of $x$ $m$/$sec^2$ horizontally and $y$ $m$/$sec^2$ vertically.

Now the resultant acceleration would be the vector sum of the individual accelerations, i.e. $vec {a_1} + vec {a_2}$.

After sometime velocity along X-direction and along Y-direction would be $xt$ $m$/$sec$ and $yt$ $m$/$sec$ respectively.

Now, in accordance with you the resultant velocity would be $sqrt{(xt)^2 + (yt)^2} = sqrt{t^2(x^2 + y^2)} = tsqrt{x^2+y^2}$.

The acceleration of the particle is $dfrac{text{Change in velocity}}{time} = sqrt{x^2 + y^2}$ $m$/$sec^2$.

You can use this understanding for any type of vector addition.

I think you mean expressing the resultant acceleration in terms of velocity similar to your approach to the resultant velocity

Let a$_x$ and a$_y$ be the accelerations along x and y axes respectively.

In time t velocity along x changes by a$_x$t and that along y changes by a$_y$t

So from what you have already found net change in v is $sqrt{{a_xt}^2+{a_yt}^2} $

So net acceleration is (change in v)/t=$sqrt{{a_x}^2+{a_y}^2} $

Problems with your approach

Your approch might be the most obvious one when starting out, however, it has some major flaws:

• You assumed constant velocities during the derivation, however, then went on to generalize it for non-constant velocities as well (this flaw can be easily overcome by replacing the $t$ in your equations by $mathrm dt$, an infinitesimal time)

• Your derivation, as it is stated, only works for Cartesian coordinates. If you try deriving the velocities using this approach in any other system of coordinates it is highly likely that your approach will fail (especially in curvilinear coordinates)

The second flaw makes your method unreliable and less useful. I advice you to adopt a better method, the one which I present below. I am only primarily showing you how to derive the acceleration, but you can (and you should) also derive the velocity the same way.

Finding acceleration

Acceleration of a particle is defined as

$$mathbf a=frac{mathrm d^2mathbf r}{mathrm dt^2}tag{1}$$

where $mathbf r$ is the position vector of the particle. Equation $(1)$ is the most general equation we could ever get. Using these equation, you can derive the correct acceleration in any coordinate system, without worrying about any angles, or using trigonometry. Moreover, this approach is fool-proof. It is quite common to not be able to "see" certain components of acceleration (see Coriolis force) when dealing with different coordinate systems, however, using equation $(1)$ you can rest assured and never worry about specific components and their validity.

In Cartesian coordinates, the equation $(1)$ simplifies to

begin{align} mathbf a&=frac{mathrm d^2(xmathbf{hat i}+ymathbf{hat j}+zmathbf{hat k})}{mathrm dt^2} &=frac{mathrm d^2(xmathbf{hat i})}{mathrm dt^2}+frac{mathrm d^2(ymathbf{hat j})}{mathrm dt^2}+frac{mathrm d^2(zmathbf{hat k})}{mathrm dt^2} &=frac{mathrm d^2 x}{mathrm dt^2}mathbf{hat i}+frac{mathrm d^2 y}{mathrm dt^2}mathbf{hat j}+frac{mathrm d^2 z}{mathrm dt^2}mathbf{hat k} end{align}

Therefore

$$|mathbf a|=sqrt{left(frac{mathrm d^2 x}{mathrm dt^2}right)^2+left(frac{mathrm d^2 y}{mathrm dt^2}right)^2+left(frac{mathrm d^2 z}{mathrm dt^2}right)^2}tag{2}$$

(In the above derivation, I pulled out the unit vectors out of the differentiation, however that might not always be the case. The unit vectors may also change with time, like they do in polar, spherical and cylindrical coordinates)

Like the way I derived the acceleration for Cartesian coordinates, you can also do so for other system of coordinates, such as polar coordinates, spherical coordinates, cylindrical coordinates and oblique coordinates. As a challenge, you could also try deriving them with your method, only to find how difficult it is to derive them using your method, and , instead, how easy it is to derive using the above method.

Starting with the general case

$$x_R=sqrt{x^2(t)+y^2(t)}tag 1$$

$Rightarrow$

$$v_R=dot{x}_R={frac {x left( t right) {frac {d}{dt}}x left( t right) +y left( t right) {frac {d}{dt}}y left( t right) }{sqrt { left( x left( t right) right) ^{2}+ left( y left( t right) right) ^{ 2}}}} tag 2$$

and $$a_R=ddot{x}_R={frac { left( {frac {d}{dt}}x left( t right) right) ^{2}+x left( t right) {frac {d^{2}}{d{t}^{2}}}x left( t right) + left( {frac {d}{dt}}y left( t right) right) ^{2}+y left( t right) {frac {d^{2}}{d{t}^{2}}}y left( t right) }{sqrt { left( x left( t right) right) ^{2}+ left( y left( t right) right) ^ {2}}}}-1/2,{frac { left( x left( t right) {frac {d}{dt}}x left( t right) +y left( t right) {frac {d}{dt}}y left( t right) right) left( 2,x left( t right) {frac {d}{dt}}x left( t right) +2,y left( t right) {frac {d}{dt}}y left( t right) right) }{ left( left( x left( t right) right) ^{2}+ left( y left( t right) right) ^{2} right) ^{3/2}}} tag 3$$

with $$x(t)=v_x,t^n~,y(t)=v_y,t^n$$ you obtain

$$v_R={t}^{n-2}nsqrt { { left( x left( t right) right) ^{2}} + left( y left( t right) right) ^{2}} $$ and

$$a_R=n,(n-1),t^{n-3},sqrt{x^2(t)+y^2(t)} $$

for $n=1$ you get your equation $$v_Rbigg|_{n=1}=frac{1}{t},sqrt{x^2(t)+y^2(t)}$$ $$v_Rbigg|_{n=2}=2,t,sqrt{x^2(t)+y^2(t)}$$

and $$a_Rbigg|_{n=1}=0$$ $$a_Rbigg|_{n=2}=frac{2}{t},sqrt{x^2(t)+y^2(t)} $$

Edit

$$boxed{a_R=frac{n-1}{t},v_R}$$