Question

An object's velocity in meters per second along a linear path is defined v(t)=t/(1+t^4), where t is greater than or equal to 0 and lesser than or equal to 3. and time is measured in seconds. Answer the following.

a) Determine the acceleration at t = 2. Include units in your answer.
b) Find the total distance traveled by the object on the interval where t is greater than or equal to 0 and lesser than or equal to 3. Include units in your answer.

Answer 1

I solved the derivative of the velocity which is -4t^2 / (t^4+1)^2. Do I plug in t=2 for part a?

Answer 2

yes that's correct

For the second part you need to find the integral of the velocity (= distance)
The find the definite integral between t=0 to t=3

Answer 3

I get a negative answer for a, though. Velocity cannot be negative, right?

Answer 4

it is not velocity you are calculating - it is acceleration.

However - both velocity and acceleration CAN be negative - it depends on the direction which you define as positive.

In this case the negative value for acceleration means that the object is slowing down at this point in its motion .i.e. the rate of change of velocity is negative

Answer 5

integral of t/ (1 + t^2) = (½) log (1 + t^2)

Answer 6

Isn't it arctan(t^2)/2?

Answer 7

because it is the integral of t/(1+t^4)

Answer 8

oh, and my integral from part a was wrong. i caught it, i actually took the derivative of 1/(1+t^4)

Answer 9

i see

Answer 10

i think i solved both parts. will you check my work?

a. Take the derivative of t/(1+t^4). That is -(3t^4 -1) / (t^4+1)^2. Plugging in t=2, I get -47/289 meters per second. This is about -.16 meters per second.

b. Take the definite integral of t/(1+t^4) which is arctan(t^2) / 2. Between the interval of 0 and 3, I would get about .73 meters per second.

Answer 11

for this integral you use the standard form INT f'(x) / f(x) = log f(x)
or rather ln f(x) - log to base e

Answer 12

0a - yes
b - (½) ln 10 - (½) ln 1 = 1.15

Answer 13

I don't understand how you are solving part b for the integral. Wouldn't you use the integration formula for integral of du/(a^2+u^2) = 1/a arctan(u/a)

Answer 14

hold on - i'm relying on memory i\ll check it out

Answer 15

http://www.wolframalpha.com/input/?i=integrate+with+respect+to+t%3A+++t+%2F+%281+%2B+t%5E2%29

- take a look

Answer 16

ah, see the given equation is actually t/1+t^4 not 1/1+t^2