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## 1. a) If a particle’s position is given by LaTeX: x\:=\:4-12t\:+\:3t^2x = 4 − 12 t + 3 t 2(where t is in seconds and x is in meters), what i

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1. a) If a particle’s position is given by LaTeX: x\:=\:4-12t\:+\:3t^2x = 4 − 12 t + 3 t 2(where t is in seconds and x is in meters), what is its velocity at LaTeX: t=1st = 1 s? b) Is it moving in the positive or negative direction of LaTeX: xx 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 calculations.) e) Is there ever an instant when the velocity is zero? If so, give the time LaTeX: tt ; if not, answer no. f) Is there a time after LaTeX: t=3st = 3 s when the particle is moving in the negative direction of LaTeX: xx? If so, give the time LaTeX: tt; if not, answer no. (Hint: Speed= LaTeX: \mid v\mid∣ v ∣)

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3 years
2021-08-02T05:09:00+00:00
2021-08-02T05:09:00+00:00 2 Answers
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## Answers ( )

Answer:a) v=-6m/s

b) negative direction

c) 6m/s

d) decreasing

e) for t=2s

f) Yes

Explanation:The particle position is given by:

a) the velocity of the particle is given by the derivative of x in time:

and for t=1s you have:

b) for t=1s you can notice that the particle is moving in the negative x direction.

c) The speed can be computed by using the formula:

d) Due to the negative value of the velocity in a) you can conclude that the speed is decreasing.

e) There is a time in which the velocity is zero. You can conclude that because if t=2 in the formula for v in a), v=0

f) after t=3s the particle will move in the negative direction, this because it is clear that 4+3t^2 does not exceed -12t.

Answer:a) v (1) = -6 m/s

b) negative x-direction

c) s ( 1 ) = 6 m/s

d) The speed decreases at t increases from 0 to 2 seconds.

e) At t = 2 s, the velocity is 0

f) No

Explanation:Given:-– The position function of the particle:

x (t) = 4 – 12t + 3t^2

Find:-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 no. (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 no.

Solution:-– The velocity function of the particle v(t) can be determined from the following definition:

v (t) = d x(t) / dt

v (t) = -12 + 6t

– Evaluate the velocity at time t = 1 s:

v (1) = -12 + 6(1)

v (1) = -6 m/s–The negative sign of the velocity at time t = 1s shows that the particle is moving in the negative x-direction.– The speed ( s ( t )is the absolute value of velocity at time t = 1s:

s ( t ) = abs ( v ( t ) )

s ( 1 ) = abs ( v ( 1 ) )

s ( 1 ) = abs ( -6 )

s ( 1 ) = 6 m/s–The speed of the particle at time t = 0,s ( t ) = abs ( -12 + 6t )

s ( 0 ) = abs (-12 + 6 (0) )

s ( 0 ) = abs ( -12 )

s ( 0 ) = 12 m/s–The speed of the particle at time t = 2,s ( t ) = abs ( -12 + 6t )

s ( 2 ) = abs (-12 + 6 (2) )

s ( 2 ) = abs ( 0 )

s ( 2 ) = 0 m/s– Hence, the speed of the particle decreases from s ( 0 ) = 12 m/s to s ( 2 ) = 0 m/s in the time interval t = 0 to t = 2 s.– As the speed decreases as time increases over the interval t = 0 , t = 2 s the velocity v(t) also approaches 0, at time t = 2 s. s ( 2 ) = 0 m/s.

– We will develop an inequality when v (t) is positive:

v (t) = -12 + 6t > 0

6t > 12

t > 2

–

So for all values of t > 2 the velocity of the particle is always positive.