# Relative velocity in two dimensional motion [closed]

I came across the following problem in my textbook:

"Two particles A and B are projected in air. A is thrown with a speed of 30m/s and B with a speed of 40m/s as shown in the figure. What is the separation between them after 1 sec?"

I approached this problem by resolving both velocity vectors into its components along the x(positive toward right) and y(positive upwards) axes. Then, I found out the velocity of A w.r.t B and solved the problem by using this relative velocity and acceleration.

In the solutions, however, they have approached it in a much quicker method as follows:

I don't understand how they've been able to find the relative velocity by taking the square root of 30²+40² directly. What point am I missing? Please do help. Thanks in advance :)

Since both have identical acceleration, that due to gravity, their relative acceleration is zero. So final answer would be the same if there were no gravity at all. Put $g=0$, orient your X-axis along direction of motion of any one particle, notice that angle between their directions of motion is $90^\circ$, and you have the quick way to arrive at final answer.
• Oh, so what if there was some definite value of relative acceleration? In that case, after orientation along one of the directions of motion, I would've had to use to $1/2at²$ term in addition to the $ut$ term to find the displacement, yes?(Where a=Relative acceleration) – user106570 Aug 22 '16 at 4:44
• If $a$ is constant, then switch to reference frame attached to any one particle. In this frame the other particle has constant acceleration $a$, and some initial relative velocity. Solve the problem with usual equations of motion. – Deep Aug 22 '16 at 4:52
this is a projectile motion problem in which there are two motions in two directions. one in the horizontal direction with uniform motion i.e. ($a_x=0$) and the other is in vertical with free falling motion i.e. ($a_y=-g$). firstly find the position the coordinates of the particles after $1\,\rm{s}$ in the $x-y$ plane and then use the Pythagorean theorem to determine the distance between them. therefore $$V_{Bx}=|V_B|\,\cos 53^\circ=(40)(0.6)=-24\,\rm{\frac m s}$$ $$V_{By}=|V_B|\,\sin 53^\circ=(40)(0.8)=32\,\rm{\frac m s}$$ in above $V_{Bx}<0$ since it lies in the negative direction of $x$ axis. $$V_{Ax}=|V_A|\,\cos 37^\circ=(30)(0.8)=24\,\rm{\frac m s}$$ $$V_{Ay}=|V_A|\,\sin 37^\circ=(30)(0.6)=18\,\rm{\frac m s}$$ Now using the kinematic relation between position and velocity in uniform and falling motions, we have uniform motion in $x$ direction: $$x_B=V_{Bx}t=\left(-24\,\rm{\frac m s}\right)(1\,\rm{s})=-24\,\rm{m}$$ $$x_A=V_{Ax}t=\left(24\,\rm{\frac m s}\right)(1\,\rm{s})=24\,\rm{m}$$ falling motion in $y$ direction: $$y_B=-\frac 1 2 gt^{2}+V_{By}t=-\frac 1 2 (10)(1)^2+(32)(1)=27\,\rm{m}$$ $$y_A=-\frac 1 2 gt^{2}+V_{Ay}t=-\frac 1 2 (10)(1)^2+(18)(1)=13\,\rm{m}$$ Using the Pythagorean theorem we can get the distance between these points in the $x-y$ coordinate as \begin{align*} d&=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}\\ &=\sqrt{\left(24-(-24)\right)^2+\left(13-27\right)^2}\\ &=\sqrt{2500}=50\,\rm{m} \end{align*}