The position of a particle moving along an x axis is given by x = 19t2 - 2.0t3,

Question

The position of a particle moving along an x axis is given by x = 19t2 - 2.0t3, where x is in meters and t is in seconds.
(a) Determine the position, velocity, and acceleration of the particle at t = 3.0 s.
x =___ m
v =____ m/s
a =___ m/s2

(b) What is the maximum positive coordinate reached by the particle?
___m
At what time is it reached?
s

(c) What is the maximum positive velocity reached by the particle?
___m/s
At what time is it reached?
s

(d) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)?
____m/s2

(e) Determine the average velocity of the particle between t = 0 and t = 3 s.
___ m/s

Explanation / Answer

a) at t= 3sec x=117 m v=38t-6t^2=60 m/s a=2 m/s^2 b) to find a maximum, differentiate the function and set the derivative to 0 . x=19t^2-2t^3 x'=38t-6t^2 this equals zero when 38t=6t^2 or when t=6.33 s when t=6.33, the particle has a position of: x=19(6.33)^2-2(6.33)^3 x=254.04 m c) the velocity is the derivative of the displacement, so the velocity is v=38t-6t^2 v is a max when dv/dt is zero, or when dv/dt=38-12t=0 this occurs when t=3.17, when v=38x3.17-6x3.17^2 =60.17 m/s d) acceleration is the derivative of velocity, or 38-12t the particle is not moving when dx/dt=0, which occurs when t=6.33; so the accel at t=6.33 is just 38-12(16.33) = -37.96 m/s^2 e) Avg velocity=avg dist/ avg time at t=0 ,x=0 at t=3 ,x= 117 Avg vel =117/3=39 m/s

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