Question

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?

Answer 1

Answer:

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

$r(t) = 10i + 15j + 100k$

The maximum values of particle will at t= 10 s

$V_{max}=10.63\ m/s$

The minimum values of particle will at t= 5 s

$V_{min}=3.6 \ m/s$

Explanation:

Given that

$r(t) = 2ti + 3tj + (100-(t- 5)^2)k$

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.

$Z=(100-(t- 5)^2)$

$\dfrac{dZ}{dt}=0-2(t-5)$

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

Position at t= 5

$r(t) = 2ti + 3tj + (100-(t- 5)^2)k$

$r(t) = 2\times 5i + 3\times 5j + (100-(5- 5)^2)k$

$r(t) = 10i + 15j + 100k$

Speed :

$V=\dfrac{dr}{dt}$

$V=2i + 3j -2(t- 5)k$

The maximum values of particle will at t= 10 s

$V_{max}=\sqrt{2^2+3^2+10^2} \ m/s$

$V_{max}=\sqrt{113} \ m/s$

$V_{max}=10.63\ m/s$

The minimum values of particle will at t= 5 s

$V_{min}=\sqrt{2^2+3^2} \ m/s$

$V_{min}=\sqrt{13} \ m/s$

$V_{min}=3.6 \ m/s$