A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 9 m,
y =6 m, and has velocity vo =(4.5 m/s)i + (−4 m/s)j. The acceleration is
given by a = (6 m/s2)i + (1 m/s2)j.
What is the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 8.5 s?
Answer in units of m
d= dinitial + vi *t+1/2 a t^2
Idont understand your question here. Just put in dinitial, vi, and a, and do it. d will be a vector
To find the displacement of the particle from the origin after 8.5 s, we can use the equations of motion.
First, let's calculate the velocity of the particle at 8.5 s using the initial velocity and acceleration:
v = vo + a * t
= (4.5 m/s)i + (-4 m/s)j + (6 m/s^2)i + (1 m/s^2)j * 8.5 s
= 4.5i - 4j + 51i + 8.5j
= 55.5i + 4.5j
Next, let's calculate the displacement of the particle by integrating the velocity with respect to time:
s = ∫v dt
= ∫(55.5i + 4.5j) dt
= (55.5t)i + (4.5t)j
Now, let's evaluate the displacement at 8.5 s:
s = (55.5 * 8.5)i + (4.5 * 8.5)j
= 471.75i + 38.25j
Finally, let's calculate the magnitude of the displacement from the origin:
|s| = √(471.75^2 + 38.25^2)
= √(222527.5625 + 1464.5625)
= √(223992.125)
≈ 473.18 m
Therefore, the magnitude of the displacement from the origin after 8.5 s is approximately 473.18 m.
To find the magnitude of the displacement from the origin after 8.5 seconds, we need to calculate the position of the particle at that time and then calculate the distance between the origin and that position.
To find the position of the particle at time t, we can use the equation of motion:
x = x0 + v0xt + (1/2)axt^2
y = y0 + v0yt + (1/2)ayt^2
Where:
x0, y0 are the initial positions (given as 9 m, 6 m).
v0x, v0y are the initial velocities (given as 4.5 m/s, -4 m/s).
ax, ay are the components of acceleration (given as 6 m/s^2, 1 m/s^2).
t is the time (given as 8.5 s).
Plugging in the values, we get:
x = 9 + (4.5)(8.5) + (1/2)(6)(8.5)^2
y = 6 + (-4)(8.5) + (1/2)(1)(8.5)^2
Simplifying the equations, we get:
x = 9 + 38.25 + 229.5
y = 6 - 34 + 36.125
x = 276.75 m
y = 8.125 m
Now, we can calculate the magnitude of the displacement from the origin using the Pythagorean theorem:
displacement = sqrt(x^2 + y^2)
displacement = sqrt((276.75)^2 + (8.125)^2)
displacement ≈ sqrt(76620.5 + 66.015625)
displacement ≈ sqrt(76686.515625)
displacement ≈ 277.075 m
Therefore, the magnitude of the displacement from the origin after 8.5 s is approximately 277.075 m.