a particle moves with the position function s = t^4 - 4t^3 - 20t^2 + 20t    t  is greater than or equal to 0at what time does the particle have a velocity of 20m/s? at what time is the...

a particle moves with the position function

s = t^4 - 4t^3 - 20t^2 + 20t    t  is greater than or equal to 0

at what time does the particle have a velocity of 20m/s?

at what time is the accelleration 0?

what is the significance of this value of t?

Asked on by ambybear

1 Answer | Add Yours

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`s(t) = t^4 - 4t^3 - 20t^2 + 20t`

The velocity function is given by the derivative of position function.

`v(t) = (ds(t))/dt`

`v(t) = 4t^3-12t^2-40t+20`

 

When the particle velocity is 20 then;

`v(t) = 20`

`4t^3-12t^2-40t+20 = 20`

`4t^3-12t^2-40t = 0`

`4t(t^2-3t-10) = 0`

`t(t^2-5t+2t-10) = 0`

`t[t(t-5)+2(t-5)] = 0`

`t(t-5)(t+2) = 0`

 

To satisfy the above equation;

`t = 0,5,-2`

 

Since `t>=0`; ` t = -2` is not an answer.

Therefore the velocity will be 20m/s when the time is 0 and 5 seconds.

 

Note:

editors are allowed to answer only one question. For the next part you can use the assistant below.

 

a(t) is the derivative of v(t).

Get the function of a(t) and find the time that a(t) = 0 by solving a(t) = 0 function. Remember `t>=0.`

 

Sources:

We’ve answered 318,988 questions. We can answer yours, too.

Ask a question