Asked by julia
A particle starts from the origin at t = 0 with an initial velocity of 5.5 m/s along the positive x axis.If the acceleration is (-2.8i + 4.1jm/s^2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate.
Answers
Answered by
drwls
The i term is the x-axis comonent and the j term is the y-axis component
Maximum x coordinate is reached when 5.5 m/s = 2.8 t
t = 1.9643 s
(a) At that time, velocity is
5.5i + (-2.8i + 4.1j)t = 4.1j*1.9643
= 8.0536j m/s (parallel to y axis)
(b) Position =
0 + (5.5 t -1.4 t^2)i + 2.05 t^2 j
Substitute 1.9643 s for t
Maximum x coordinate is reached when 5.5 m/s = 2.8 t
t = 1.9643 s
(a) At that time, velocity is
5.5i + (-2.8i + 4.1j)t = 4.1j*1.9643
= 8.0536j m/s (parallel to y axis)
(b) Position =
0 + (5.5 t -1.4 t^2)i + 2.05 t^2 j
Substitute 1.9643 s for t
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