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?
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`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.
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.`
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