A particle\'s position is given by x = 9.00 - 15.00t + 3t2, in which x is in met

Question

A particle's position is given by x = 9.00 - 15.00t + 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". How do you solve this question? Could you please show me the steps that you took to get this answer.

Explanation / Answer

x = 9.0 - 15.0t + 3t^2

(a) v = dx/dt = 0 - 15.0 + 6t = 6t - 15.0

at t = 1 s, v(1) = 6*1 - 15 = -9 m/s

(b) at t = 1 s

x(1) = 9.0 - 15*1 + 3*1^2 = -3 m

So, the particle is moving in negative direction of x.

(c) Speed = 9 m/s.

(d) acceleration, a = dv/dt = 6 m/s^2

So the speed is increasing.

(e) suppose at time, t, velocity = 0

therefore, v = 6t - 15 = 0

=> t = 15/6 = 2.5 s

(f) Put t = 3 s in the displacement expression -

x(t) = 9 - 15*3 + 3*3^2 = 9 - 45 + 27 = -9 m

So the particle is moving is negative x direction.

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