A particle\'s position is given by x = 16.0 - 12.00t + 3t2, in which x is in met

Question

A particle's position is given by x = 16.0 - 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".

Explanation / Answer

x = 16.0 - 12.00t + 3t^2 v = -12 +6t at t=1 v = -6m/s negative direction its speed is 6m/s speed is increasing just thn yes v = 0 at t = 2 no after t = 3 , v is always positive and it moves in positive after that , answer 0

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