Question

A partivle moves along the x-axis, its position at time t given by x(t)=t/(1+t^2), t>=0, where t is measured in seconds and x in meters.
a) find the acceleration at time t. When is it 0?
b) Graph the position, velocity, and acceleration functions for 0<=t<=4.
c) When is the particle speeding up? When is it slowing down?

Answer 1

you got \[x(t)=\frac{ t }{ 1+t ^{2} }\]
a) In order to find the acceleration, take the second derivative then plug in t=0.
c) Particle is speeding up when acceleration is positive, particle is slowing down when acceleration is negative.

Answer 2

Thank you! would the answer be just 0 for a)?

Answer 3

I get 0 for a... here's what I get:
\[x'(t) = \frac{ -2t ^{2} }{ (1+t ^{2})^{2} }+\frac{ 1 }{ 1+t ^{2} }\]
\[x''(t) = \frac{ -8t ^{3} }{ (1+t ^{2})^{3} }+\frac{ -4t }{ (1+t)^{2} }-\frac{ 2t }{ (1+t ^{2})^{2} }\]
\[x''(0)=0\]

Answer 4

How do you find c?

Answer 5

you can find C from the graph of x'(t) and x''(t)

Answer 6

Thank you soo much!