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A particle's position is given byx = 5.00 - 12.00t + 3t2 , in which x is in meters and t is in seconds. (a) What is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer "0". (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer "0".

Abenazer , 7 Years ago
Grade 12th pass
anser 1 Answers
Vikas TU

Last Activity: 7 Years ago

Velocity at t=1 sec,
v = dx/dt = 0 (check first)
-12 + 6t = 0
t = 2 sec.
In one second, v = -12 + 6 => -6 m/s that is  in opposite direction.
speed is the magnitude hence it is 6 m/s
At t= 2 sec,
v = 0
Therfore the speed is decreasing.

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