The position of a particle moving along an x axis is given by x = 10 t 2 ? 6.0 t

Question

The position of a particle moving along an x axis is given by x = 10t2 ? 6.0t3, where x is in meters and t is in seconds.

(a) Determine the position of the particle at t = 3.0 s.
_________ m

(b) Determine the velocity of the particle at t = 3.0 s. (Indicate the direction with the sign of your answer.)
__________ m/s

(c) Determine the acceleration of the particle at t = 3.0 s. (Indicate the direction with the sign of your answer.)
___________m/s2

(d) What is the maximum positive coordinate reached by the particle?
___________ m

(e) At what time is it reached?
___________s

(f) What is the maximum positive velocity reached by the particle?
____________ m/s

(g) At what time is it reached?
__________ s

(h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (Indicate the direction with the sign of your answer.)
____________ m/s2

(i) Determine the average velocity of the particle between t = 0 and t = 3 s. (Indicate the direction with the sign of your answer.)
____________ m/s

Explanation / Answer

x = 10*3^2 - 6*3^3= -72 m

b) v = dx/dt = 20 t - 18 t^2 = 20*3 - 18*3^2= -102 m/s

c) a = dv/dt = 20-36*t = 20-36*3= -88 m/s^2

d) max when v = 0

20 t - 18t^2 = 0

t=0 or t = 1.11

so x = 10*1.11^2 - 6*1.11^3= 4.12 m

e) t = 1.11 s

f) max v when a = 0

20 - 36*t = t

t = 20/36

v = 20*20/36 - 18*(20/36)^2= 5.56 m/s

g) t = 20/36 = 0.56 s

h) when v = 0

t = 1.11
a = 20 -36*1.11=-19.96 m/s^2

i) average v = dx/dt=-72/3 = -24 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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