The position of a particle moving along a straight line is given by s(t) = -t/t^

ID: 2828560 • Letter: T

Question

The position of a particle moving along a straight line is given by

s(t) = -t/t^2+1

(a) Find v(t), the velocity of the particle at time t, and determine when the particle is at rest.

b. When is the particle moving forward (i.e. in the positive direction) and when is the particle

moving backward (i.e. in the negative direction)? Write your answer in interval notation.

c. Where is the particle when it is at rest? Using the line below, sketch motion of the particle

from time t = 0 to time t = 2.

d. What is the total distance traveled by the particle at time t = 2?

Given ,

s(t) = -t/t^2+1

a) The velocity of the particle is given by v = d/dt (s(t))

= d/dt (-t/t^2+1)

= [ 2t(-t) - (t^2+1) (-1) ] / (-t/t^2+1)^2

= (-t^2 + 1 / t^2) (t^2+1)^2

= (1-t^4) (1+t^2) / t^2

when the particle is at rest v=0

i.e (1-t^4) (1+t^2) / t^2 = 0

(1-t^4) = 0

implies t = 1

Hence at t= 1 , the particle is at rest

b) The particle is moving forward when (1-t^4) > 0

i.e t^4 < 1

-1 < t < 1

when the paricle is moving backward (1-t^4) < 0

i.e t^4 > 1

t lies in (-infinity , -1) U (1,infinity)

c) The particle is at rest at t= 1

therefore s(1) = - 1/2

at t=0 , s(o) = infinity , s(2) = -2/5

The path is like a straight line touching x axis at t=1

and having negative y axis values after that till y - -2/5 at x=2

d) distance = velocity * time = v(2) * 2

= -38.75 * 2

= 77.5 units

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