2D Displacement:
Particle moving, V(naught) 6.5 m/s(i-hat), -2.5 m/s(j-hat) arrow above V(naught) pointing in positive x direction.
Also a= 6.5m/s^2 (i-hat), 1 m/s^2 (j-hat)
I have V(sub-x) of 65 & V(sub-y) of 6.5, both after 9 seconds
w/ initial point (4.5, 6.5). I need to know the displacement from the origin (0,0), also after 9 seconds.
The answer given is 327.169 m.
I'm not sure what formula or process they used to get there, I tried drawing my vectors "head to tail" but then I noticed that after 9 seconds the y-coordinate in the initial point minus v(sub-y) is equal to y=0 which would eliminate the y-component. Don't know where I went wrong, but that's why I'm here.
To find the displacement from the origin (0,0) after 9 seconds, we can break down the problem into components and then use vector addition.
First, let's find the final position of the particle in the x-component. We know the initial velocity V₀x = 6.5 m/s and the acceleration ax = 6.5 m/s². We can use the formula for displacement in the x-direction:
x = V₀x * t + (1/2) * ax * t²
where t is the time. Plugging in the values, we get:
x = 6.5 m/s * 9 s + (1/2) * 6.5 m/s² * (9 s)²
= 58.5 m + 263.25 m
= 321.75 m
Now, let's find the final position of the particle in the y-component. We know the initial velocity V₀y = -2.5 m/s and the acceleration ay = 1 m/s². Again, we use the formula for displacement in the y-direction:
y = V₀y * t + (1/2) * ay * t²
Plugging in the values, we get:
y = -2.5 m/s * 9 s + (1/2) * 1 m/s² * (9 s)²
= -22.5 m + 40.5 m
= 18 m
Next, we can find the displacement from the origin by using the Pythagorean theorem. The displacement (D) is given by:
D = √(x² + y²)
Plugging in the values we found:
D = √(321.75 m)² + (18 m)²
= √(103518.0625 m² + 324 m²)
= √(103518.0625 m² + 104976 m²)
= √(208494.0625 m²)
= 455.9 m
Therefore, the displacement from the origin after 9 seconds is approximately 455.9 m. It seems there was a mistake in the answer given (327.169 m).