What is the Velocity and Displacement of a Particle After 2.0s?

AI Thread Summary
The particle's initial velocity is -1.0 m/s in the x-direction and 3.0 m/s in the y-direction, with accelerations of 0.5 m/s² in the x-direction and -0.9 m/s² in the y-direction. After 2.0 seconds, the velocity is calculated to be 0 m/s in the x-direction and 1.2 m/s in the y-direction. The displacement in the x-direction is -1.0 m, while the displacement in the y-direction is 4.2 m. Therefore, the final position of the particle after 2.0 seconds is -1.0 m in the x-direction and 4.2 m in the y-direction. The calculations confirm the particle's motion under the given conditions.
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Homework Statement



A particle's initial velocity is given by v0x = -1.0m/s and v0y = 3.0m/s; it's acceleration is ax = 0.50m/s2 and ay = -0.90m/s2. (a) Calculate its velocity 2.0s later. (b) Calculate its displacement at that time.

Homework Equations



vx = v0x + axt
∆x = v0xt + 1/2axt2
∆y = v0yt + 1/2ayt2

The Attempt at a Solution



v = -1.0m/s(x hat) + 3.0m/s(y hat)
a = 0.5m/s2(x hat) - 0.9m/s2(y hat)

Solve for velocity at 2.0s
vx = v0x + axt
vx = (-1m/s) + (0.5m/s2)(2.0s)
vx = 0m/s

vy = v0y + ayt
vx = (3m/s) + (-0.9m/s2)(2.0s)
vx = 1.2m/s

v = 0m/s(x hat) + 1.2m/s(y hat)

Calculate displacement.
∆x = v0xt + 1/2axt2
∆x = (-1m/s)(2.0s) + 1/2(0.5m/s2)(2.0s)2
∆x = -1m

∆y = v0xt + 1/2axt2
∆y = (3m/s)(2.0s) + 1/2(-0.9m/s2)(2.0s)2
∆y = 4.2m

y = -1.0m(x hat) + 4.2m(y hat)
 
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