The position of a particle moving along an x axis is given by x = 23t2 5t3, wher

Question

The position of a particle moving along an x axis is given by x = 23t2 5t3, where x is in meters and t is in seconds.

(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(t) = 23t^2 - 5t^3

velocity = dx/dt = 46t - 10t^2

for maxium velocity dx/dt = 0

since dx/dt = 46 - 20t = 0

=> t = 2.3 s

f) maximum positive velocity = 46(2.3) - 10(2.3^2) = 52.9 m/s

g) it is reached at t = 2.3 s

h) particle is not moving => velocity = 0

v = 46t - 10t^2 = 0

=> t =0 ,t= 4.6 s

acceleration = dv/dt = 46 -20t

at t = 4.6 s

acceleration = 46 - 20*4.6 = -46 m/s^2   

particle is deaccelerating.

i)t=0 t0 t=3 s

average velocity = total distance travelled /total time = [x(3)-x(0)]/3 = (72-0)/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.