- #1
RoganSarine
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"Simple" Coordinate Problem
The following happens on a 2D X-Y Plane.
A particle accelerates at {3t m/s2}i + 4t m/s2}j where t = seconds
At t = 0, the position of the particle is {20.0 m}i + {40.0m}j
At t = 0, the velocity of the particle is {5.00 m/s}i + {2.00 m/s}j
At t = 4,
(a) What is the position vector in unit-vector notation
(b) What is the angle between its direction of travel and the positive direction of the x-axis
dx = Vox*t + .5(a)t^2
Dy = Voy*1 + .5(a)t^2
dx = Vox*t + .5(a)t^2
dx = 5(4) + .5(3(4)) (4)^2
dx = 20 + 6(16)
dx = 116
Dy = Voy*1 + .5(a)t^2
Dy = 2(4) + .5(4(4)) (4)^2
Dy = 8 + 8(16)
Dy = 136
Therefore, it's position vector should be
P = {136 m}i + {176 m}j
If we remember to add on the initial displacement.
The answer is {72.0 m)i + {90.7 m}j
Homework Statement
The following happens on a 2D X-Y Plane.
A particle accelerates at {3t m/s2}i + 4t m/s2}j where t = seconds
At t = 0, the position of the particle is {20.0 m}i + {40.0m}j
At t = 0, the velocity of the particle is {5.00 m/s}i + {2.00 m/s}j
At t = 4,
(a) What is the position vector in unit-vector notation
(b) What is the angle between its direction of travel and the positive direction of the x-axis
Homework Equations
dx = Vox*t + .5(a)t^2
Dy = Voy*1 + .5(a)t^2
The Attempt at a Solution
dx = Vox*t + .5(a)t^2
dx = 5(4) + .5(3(4)) (4)^2
dx = 20 + 6(16)
dx = 116
Dy = Voy*1 + .5(a)t^2
Dy = 2(4) + .5(4(4)) (4)^2
Dy = 8 + 8(16)
Dy = 136
Therefore, it's position vector should be
P = {136 m}i + {176 m}j
If we remember to add on the initial displacement.
The answer is {72.0 m)i + {90.7 m}j