Respuesta :
Answer:
The speed is minimum at t = 2.
Step-by-step explanation:
The position of a particle is given by [tex]r(t) = t^{2}, 2t, t^{2} - 16t.[/tex]
Differentiating the above with respect to t, we get, velocity v(t) = [tex][2t, 2, 2t - 16][/tex].
Hence, the speed of the particle at t is [tex]\sqrt{(2t)^2 + (2)^2 + (2t - 16)^2 } = \sqrt{4t^2 + 4 + 4t^2 - 64t + 256} = \sqrt{16t^2 -64t + 260}[/tex].
The speed will be minimum, if
[tex]\frac{d \sqrt{16t^2 - 64t + 260} }{dt} = 0\\\frac{32t - 64}{2\sqrt{16t^2 - 64t + 260}} = 0\\\frac{16t - 32}{\sqrt{16t^2 - 64t + 260}} = 0\\t = \frac{32}{16} = 2[/tex].
The speed will be minimum at t = 2.
At t=2, the speed is 14.
The speed is minimum at t = 4 seconds.
The position function of a particle is given by
[tex]r(t) = t^{2},2t,t^{2}-16t[/tex]
The velocity is given as,
[tex]v=\frac{dr(t)}{dt} =2t,2,2t-16[/tex]
Speed is computed as,
[tex]Speed=\sqrt{(2t)^{2}+2^{2}+(2t-16)^{2} }\\ \\v=\sqrt{8t^{2}-64t+260 }[/tex]
Differentiate above expression and equate to zero.
[tex]\frac{dv}{dt}=\frac{16t-64}{2\sqrt{8t^{2} -64t+260} } =0\\\\16t=64\\\\t=64/16=4[/tex]
Thus, speed is minimum at t = 4 seconds.
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