If a particle is travelling at a velocity of $v = (11 \hat{i} + 10\hat{j}) \frac m s$, and undergoing constant acceleration $a = (-1.20\hat{i} + 0.30\hat{j}) \frac m {s^2}$, in what direction is it moving? I have been trying all day to get this. I've worked out that the particle crosses the $y$-axis ($x=0$) when $t = 18s$ with the $y$-coordinate at that time being $y=230m$, travelling $19 \frac m s$... I just can't work out what direction it it is moving (anticlockwise from the positive $x$-axis).
2026-03-25 19:25:10.1774466710
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Particles in motion
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x=11t+C Since x=0 @ t=18 => 0=11*18+C => C=-198 => x=11t-198
y=10t+C1 Since y=230 @ t=18 => 230=10*18+C1 => C1=50 => y=10t+50
θ(anticlockwise)= arctan2 ((10t+50)/(11t-198)) Let t=0 => approximately 143 deg
θ(anticlockwise)= arctan2 ((10t+50)/(11t-198)) Let t=1 => approximately 139 deg https://en.wikipedia.org/wiki/Atan2
Therefore, the direction would be clockwise.
I am confused on how the object can travel at 19 m/s at y=230m when your velocity in the y direction is at 10 m/s. I am having a hard time understanding the problem in that case. Hope this helps out enough with the problem.
If the initial velocity is $v = (11 \hat{i} + 10\hat{j}) \frac m s$, and the acceleration $a = (-1.20\hat{i} + 0.30\hat{j}) \frac m {s^2}$. Then, the velocity dependent on time is $$v=v_0+at=((11-1.20t)\hat{i},(10+0.3t)\hat{j})$$ The direction will depend then on time. For example, the angle with respect to the $x-axis$ can be determined as $$\theta=\mbox{atan}\left(\frac{v_y}{v_x}\right)=\mbox{atan}\left(\frac{10+0.3t}{11-1.20t}\right)$$ I hope it will be useful!