A particle\'s position is given by x = 13.0 - 12.00t + 3t2, in which x is in met

Question

A particle's position is given by x = 13.0 - 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

I presume this is for a calc based course:

a) v = dx/dt = -12+6t; when t=1 v = -6m/s

b) since v<0, the motion is in the negative direction

c) speed is 6m/s

d) to determine whether the speed is increasing or decreasing, take the derivative of v: dv/dt = 6, sincd dv/dt>0, the velocity is increasing

e) v= -12 + 6t which is equal to zero when t = 2s

f) for t>3, v is always > 0 since 6t> 12 for all t > 2

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