a particle moves along a given path r=2ti+3tj+(100-(t-5)^2)k for time 0=<t=>10 w

Question

a particle moves along a given path r=2ti+3tj+(100-(t-5)^2)k for time 0=<t=>10

when is the particle at its highest point?

when is the particle moving the fastest?

when is the particle moving the slowest?

Explanation / Answer

a> The particle is at highest point when its co,ordinate along k is maximum. i.e. 100-(t-5)^2 is maximum which is at t = 5. b> v(t) = 2 i + 3 j - 2*(t-5) k |v(t)| = sqrt( 13+4*(t-5)^2) |v(t)| is maximum or minimum when (t-5)^2 is maximum or minimum. Thus the particle is fastest at t = 0 or t = 10 c> The particle is slowest at t = 5.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.