Question

A particle moves along the x-axis so that its velocity v at time t greater than or equal to 0 is given by v(t)=sin(t^2). The graph of v for 0 less than or equal to t less than or equal to square root 5pi. THe position of the particle at time t is x(t) and its position at time t=0 is x(0)=5 a.) find the acceleration of the particle at time t=3. b.) find the total distance traveled by the particle from time t=0 to t=3 c.) find the position of the particle at time t=3 d.) for 0 is less than or equal to t is less than or equal to square root of 5pie, find the rate at which the particle is the farthest to the right. explain your answer.

Answer 1

The acceleration \(a(t)=\frac d{dt} v(t)\).

Answer 2

So \(a(3) = \frac d{dt} v(t) ,t=3\).

Answer 3

for b), we us: \[
\Delta x = x(3)-x(0) = \int_0^3v(t)~dt
\]Because \(v(t) = \frac d{dt}x(t)\).

Answer 4

Will, this would be the net distance traveled.

Answer 5

For the last one, we can just use \(x(0) = 5\) \[
x(3) - 5 = \int_0^3v(t)~dt
\]

Answer 6

I just added part d.) of the problem, if you could help me with it that'd be great. thanks for the help so far!