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A particle experiences a constant acceleration that is south at 2.50 meters per second squared. At t=0, its velocity is 40.0 meters per second east. What is its velocity at t=8.00 seconds?
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Answered by
MathMate
The acceleration is due south, so the east-west component (40 m/s) is not altered.
For the north-south component, at t=0,
initial velocity = 0 m/s
after 8 seconds,
velocity v=0+(-2.5)8 = -20 m/s (<0 for south)
Velocity = (40,-20) m/s
Magnitude = √(40²+20²)
=44.72 m/s
direction:
atan(-20/40)=-26.57°, or
26.57° south of east.
For the north-south component, at t=0,
initial velocity = 0 m/s
after 8 seconds,
velocity v=0+(-2.5)8 = -20 m/s (<0 for south)
Velocity = (40,-20) m/s
Magnitude = √(40²+20²)
=44.72 m/s
direction:
atan(-20/40)=-26.57°, or
26.57° south of east.
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