A particle\'s position is given by x = 1.00 - 15.00t + 3t2, in which x is in met

Question

A particle's position is given by x = 1.00 - 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".

Explanation / Answer

Assuming its x = 1.0 - 15.00t + 3t^2 Metres (m)

Now velocity is metres per second = m/s it is a rate of CHANGE of distance (dx/dt)
Your equation needs to be differentiated to get velocity

since dx/dt has units m/s (think of this as metres divided by seconds)

Lets do it

V = Velocity = dx/dt = 0 - 15 +3*2t = -15 +6t m/s

just put t=1
we get V = -15+6 = -11 m/s (a)
(minus just means it happens to be moving to the left (negative) at this point assuming positive means to the right) (b)

Now velocity and speed are the same EXCEPT if you say velocity you must also specify a direction
So the speed is just 11 m/s (drop any sign + or 1) (c)


Now acceleration is the rate of CHANGE of velocity (dv/dt) so we must differiate yet again


V = = -15 +6t

A = Acceleration = dv/dt = +6 m/s^2 (note units are m divide by s twice!)

SO

the rate of change of velocity is +6 m/s^2

Since this is positive the particle has constant acceleration of +6
so INCREASING speed (d)


Question f

We must consider t > 3 s

We know x = 1.0 - 15.00t + 3t^2

X will be negative whenever - 15t + 3t^2 exceeds 1

when t=3 x = 1 - 45 + 27 = -17
when t=4 x = 0 and increases there after because 3t^2 increasing rapidly.

so answer is no (f)

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