Answer to Question #143406 in Civil and Environmental Engineering for Becky Thomas

Question #143406
The position of a particle which moves along a straight line is defined by the relation x=t^3-6t^2-15t+40, where x is expressed in feet and t in second.A. determine the time at which the velocity will be zero B.the position and distance traveled by particle at that time C.the acceleration of the particle at that time D.the distance traveled by the particle from t=4s to t=6s.
1
Expert's answer
2020-11-18T13:58:36-0500

A. Velocity of the particle is "v(t) = x'(t) = 3 t^2 - 12 t - 15".

Hence "v(t) = 0 \\Rightarrow 3 t^2 - 12 t - 15 = 0", which has positive solution "t = 5", so velocity is zero at "t = 5".

B. Substituting "t = 5" into "x(t)", obtain "x(5) = -60". The distance is "|x(5) - x(0)| = |-60 - 40| = 100".

C. Acceleration is "a(t) = v'(t) = 6 t - 12", hence "a(5) = 18".

D. "d(4, 6) = d(4,5) + d(5,6)", since the body has stopped at "t=5", and started moving into opposite direction.

"d(4,5) = |-52 - (-60)| = 8",

"d(5,6) = |-50 - (-60)| = 10",

hence "d(4,6) = 8 + 10 = 18"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS