A particle moves along the x-axis, its position at time t given by x(t) = t/(1+t^2), t > or equal to 0?

Question

a) When is the particle moving to the right, when is it moving to the left?

b) Find the total distance traveled during the first 4 seconds

c) find the acceleration time at t. When is it 0?

d) When is the particle speeding up? When is it slowing down?


I have a few problems like this and any help that can be offered on any part is appreciated.

Answer 1

James' answer is incorrect.

jmurphy/odu83 is on the right track, but got the wrong expression for the second derivative (the acceleration), and he uses the wrong logic for determining when the particle is speeding up or slowing down (and his answer is therefore incorrect, and would be so even if his expression for the acceleration were correct).

a) The particle is moving to the right whenever the velocity is positive, and moving to the left whenever the velocity is negative. The velocity of the particle is given by the derivative of x(t) with respect to time:

v(t) = dx(t)/dt = (1-t^2)/((1+t^2)^2)

which is the same expression as odu83 obtained. The only positive value of t for which this expression is equal to zero is t = 1. For 0<=t < 1, the velocity is positive, and the particle is moving to the right. At t = 1, the particle is momentarily at rest (reversing direction), and for t > 1, the velocity is negative, and the particle is moving to the left.

b) For part (b), you can do as odu83 has done, and integrate the absolute value of the velocity, but there is a simpler way. We know that from t = 0 to t = 1, the particle is always moving to the right, reaching a maximum displacement of x = 0.5 at t = 1 (found by simply plugging t = 1 into the expression for x(t)). Then, from t = 1 to t = 4, the particle moves from x = 0.5 to x = 0.235, or a distance of (0.5 - 0.235) = 0.265, so the total distance is the sum of 0.5 + 0.265 = 0.765

(c), the acceleration is given by the second derivative of x(t) with respect to time. The correct expression for this is:

a(t) = dv(t)/dt = 2*t*(t^2 - 3)/(1 + t^2)^3


The acceleration is zero at t = 0 and again at t = sqrt(3). The acceleration is negative for 0 < t < sqrt(3), and positive for t > sqrt(3).


(d) The particle will be slowing down whenever the acceleration and velocity have opposite signs, and speeding up whenever the acceleration and velocity have the same sign (i.e, slowing when the acceleration and velocity vectors are opposed, and increasing speed when they are aligned).

Summarizing the signs of v and a for various ranges of t:

0 < t < 1, v>0, a<0, (Particle slowing)

1 < t < sqrt(3), v< 0, a<0, (Particle speeding up)

sqrt(3) < t, v<0, a>0, (Particle slowing down)

Answer 2

James is not correct.

a) The particle moves to the right for 0<=t<=1 and then to the left for 1<=t<= infinity in the limit x-> infinity it reaches x=0 again

b)James gave the position at t=4, not the total distance traveled. If you look at t=1, the particle is at 0.5. For t>1, the particle retreats and at 4 it is located at 4/17, or 0.235.

The distance is found as the integral of the speed from one point to another.

Speed is the first derivative, or
v(t)=(1-t^2)/
(1+2*t^2+t^4)

This must be integrated first between 0 and 1, and then integrated again between 1 and 4 since the speed becomes negative at t=1

for 0<=t<=1 distance is .5
for 1<=t<=4 distance is 4/17-.5 or -.265
so the total distance is
.5+.265=.765

c) Acceleration is the second derivative
a(t)=(-6*t-8*t^3-2*t^5)/
(1+2*t^2+t^4)^2

it has negative magnitude for all values of t>0
so the particle is always slowing down. The acceleration hits a minima at t=0.5, when the negative magnitude reaches -1.664

j

Answer 3

a)t>0
b)4/17
c)x''(t=0)
d)x''(t)>0 ; X''(t)<0