The position of a particle is given by s=t^3-7.5t^2-4t,when does the particle reach a velocity of 104m/s?

Answer 1

s=t^3-7.5t^2-4t
ds/dt=3t^2-15t-4
equating ds/dt to 104
3t^2-15t-4=104
=>3t^2-15t-108=0
dividing throughout by 3
t^2-5t-36=0
factoring
(t-9)(t+4)=0
=.t=9 0r t=-4
ignoring the negative value as it is nor permissible the particle will reach a vel of 104 m/s when t=9sec

Answer 2

first of all, you differentiate the equation to get the equation of the velocity. so v will be ds/dt= 3(t^2) - 15t -4
so v = 3(t^2) - 15t -4 = 104
3(t^2) -15t -108 = 0 divide all by 3
(t^2) -5t -36 = 0 (t-9) (t+4) = 0
so t = 9 sec. or -4 sec. but since there are no seconds in minus so the particle reaches 104m/s after 9 seconds.

Answer 3

The velocity function is found by taking the derivative of the position function s = t^3 - 7.5t^2 - 4t

v = s' = 3t^2 - 15t - 4

Velocity = 104 m/s = 3t^2 - 15t - 4

Do some algebra to solve for the time:

3t^2 - 15t -108 = 0

(3t + 12)(t - 9) = 0

3t + 12 = 0 or t - 9 = 0
t = -4sec or 9sec

Since time cannot be negative, 9 sec is your answer.

Answer 4

If s=t^3-7.5t^2-4t then find v(t) by differentiating:
v(t) = 3t^2 - 15t - 4
And you want v = 104, so 3t^2 - 15t - 4 = 104
Then so 3t^2 - 15t - 108 = 0
t^2 - 5t - 36 = 0
(t + 4)(t - 9) = 0
t = -4 is not allowed (t<0)
So at t = 9s the speed is 104 m/s

Answer 5

It will reach a velocity of 104m/s after 9 secs.

Answer 6

depends what is the initial velocity of the particle

v(t) = s'(t) + v(0) = 3t^2 -15t -4 + v(0)

3t^2 -15t -4 + v(0) - 104 = 0;
solve t .