The origin is on flat ground, and the z-axis points upward. For time 0 ≤ t ≤ 10, a particle moves along a path given by ⃗r(t) = 2ti + 3tj + (100 − (t − 5)2 ) ⃗k. For what time(s) 0 ≤ t ≤ 10 is the particle at its highest point and what is its position then? For what time(s) 0 ≤ t ≤ 10 is the particle moving fastest? What is the fastest speed? For what time(s) 0 ≤ t ≤ 10 is the particle moving slowest? What is the slowest speed?

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Netta00 Netta00 · Answer 1 · 2019-10-05T18:48:49+02:00

Answer:

t= 5 s the particle will be on the highest position.

[tex]r(t) = 10i + 15j + 100k[/tex]

The maximum values of particle will at t= 10 s

[tex]V_{max}=10.63\ m/s[/tex]

The minimum values of particle will at t= 5 s

[tex]V_{min}=3.6 \ m/s[/tex]

Explanation:

Given that

[tex]r(t) = 2ti + 3tj + (100-(t- 5)^2)k[/tex]

Time when the particle will be on the highest position:

When the Z component of r(t) will be maximum then the particle will be on the highest position.

[tex]Z=(100-(t- 5)^2)[/tex]

[tex]\dfrac{dZ}{dt}=0-2(t-5)[/tex]

It means at t= 5 s the particle will be on the highest position.

Position at t= 5

[tex]r(t) = 2ti + 3tj + (100-(t- 5)^2)k[/tex]

[tex]r(t) = 2\times 5i + 3\times 5j + (100-(5- 5)^2)k[/tex]

[tex]r(t) = 10i + 15j + 100k[/tex]

Speed :

[tex]V=\dfrac{dr}{dt}[/tex]

[tex]V=2i + 3j -2(t- 5)k[/tex]

The maximum values of particle will at t= 10 s

[tex]V_{max}=\sqrt{2^2+3^2+10^2} \ m/s[/tex]

[tex]V_{max}=\sqrt{113} \ m/s[/tex]

[tex]V_{max}=10.63\ m/s[/tex]

The minimum values of particle will at t= 5 s

[tex]V_{min}=\sqrt{2^2+3^2} \ m/s[/tex]

[tex]V_{min}=\sqrt{13} \ m/s[/tex]

[tex]V_{min}=3.6 \ m/s[/tex]