# Thread: Don't get this velocity question?

1. ## Don't get this velocity question?

Particle moves along the x-axis so that its velocity at time t greater or equal to 0 is given by $v(t) = sin(t^2)$

Graph shown for 0 to sqrt(5pi). Position of the particle @ time t is x(t) and its position at time t = 0 is x(0) = 5

a) Find the acceleration of the particle at the time t = 3

I got the answer as 0.. IS this right??

The other ones I don't get,,,

b) Find the total distance traveled by the particle from time t = 0 to t= 3..

Do I take the integral of $sin(t^2)$??? from 0 to 3??

c) FInd the postion of the particle at time t = 3

d) for 0 to sqrt(5pi) with t in between.. find the time t at which the particle is farthest to the right. Explain your answer.

2. Originally Posted by elpermic
Particle moves along the x-axis so that its velocity at time t greater or equal to 0 is given by $v(t) = sin(t^2)$

Graph shown for 0 to sqrt(5pi). Position of the particle @ time t is x(t) and its position at time t = 0 is x(0) = 5

a) Find the acceleration of the particle at the time t = 3

I got the answer as 0.. IS this right??

The other ones I don't get,,,

b) Find the total distance traveled by the particle from time t = 0 to t= 3..

Do I take the integral of $sin(t^2)$??? from 0 to 3??

c) FInd the postion of the particle at time t = 3

d) for 0 to sqrt(5pi) with t in between.. find the time t at which the particle is farthest to the right. Explain your answer.
first of all, note that this problem requires the use of a graphing calculator.

a. $a(3) = v'(3) = -5.467$ ... use the nDeriv function on your calculator.

b. total distance = $\int_0^3 |v(t)| \, dt = 1.702$ ... use the fnInt function on your calculator.

c. $x(3) = x(0) + \int_0^3 v(t) \, dt = 5.774$ ... again, use fnInt

d. looking at the graph, the particle will be farthest to the right at the first time $v(t) = 0$ , at $t = 1.772$