A particle moves along the x-axis with velocity at time t is greater than or equal to 0?

Question

given by v(t) = -1 + e^(1-t).
(a) Find the acceleration of the particle at time t = 3
(b) Is the speed of the particle increasing at time t = 3? Give a reason for your answer.
(c) Find all values of t at which the particle changes direction. Justify your answer.
(d) Find the total distance traveled by the particle over the time interval 0 is less than or equal to t is less than or equal to 3.

Answer 1

A) To find a from v, take a derivative with respect to time. That'll give you an equation for a(t).
B) If a(3) is positive, then the speed is increasing, otherwise it's negative.
C) Set a(t) = 0 and find all of those points.
D) Integrate v(t) from 0 to 3.

I'm not plugging anything in, because you should be able to figure it out from that.

Answer 2

a(t) = - e^(1 - t)
(a) a(3) = - e^-2 ≈ - 0.1353353
(b) v(3) = - 1 + e^-2 = -0.8646647
The speed is increasing because the acceleration is in the same direction as the velocity.
(c) When v = 0
e^(1-t) = 1
t - t = 0
t = 1
(d)
s = - t - e^(1-t) + C
s = |[- 3 - 0.1353353] - [-1 - 1]| + |[- 1 - 1] - [0 - e]|
s = |[- 3.1353353 + 2| + |[- 2 + e]|
s = |- 1.1353353| + |0.7182818|
s = 1.853617

Answer 3

a(t) = 2 - 5(pi) cos x and x(t) = t^2 +( 5/(pi)) cos t to detect a (t) you're taking the spinoff for x(t) you're taking the intergral. Graph velocity and each time it truly is decrease than 0 on the graph it truly is shifting to the left. stable good fortune. desire that enables... sorry regardless of if it truly is puzzling , i attempted. :)