Applications Of Calculus, Help Please? (1 Viewer)

[LuckyLuke007]

Solved

 

[elseany]

a) The particle is at rest when its velocity is zero, so we must find when 1 + 2cost = 0

1 + 2cost = 0
cost = -1/2
now the first two times this will happen is when t = 2pi/3 and 4pi/3 (draw a graph to help you)

b)We first have to integrate the expression for its velocity with respect to time to get an equation for the particles displacement.

so we know that v = x' = 1 + 2cost
so x = t - 2sint + c
we know that at the start of the particles motion it is at the origin, this is telling us that when t = 0 x = 0, so we sub in these numbers to evaluate c and we get c = 0
so x = t - 2sint

now to find the distance between the two points, we evaluate the distance to each point and subtract them.
so at x₁; t = 2pi/3
x₁ = 2pi/3 - sqrt(3)
and at x₂; t = 4pi/3
x₂ = 4pi/3 + sqrt(3)
distance = x₂ - x₁
= 4pi/3 + sqrt(3) - (2pi/3 - sqrt(3))
= 2pi/3 + 2sqrt(3) meters

 

[LuckyLuke007]

Thanks a bundle <!-- google_ad_section_start(weight=ignore) -->elseany<!-- google_ad_section_end --><SCRIPT type=text/javascript> vbmenu_register("postmenu_2916388", true); </SCRIPT>