The position of a particle moving along an axis is given by x = 22t^2 where x is

Question

The position of a particle moving along an axis is given by x = 22t^2 where x is in meters and t is in seconds. Determine the position of the particle at t = 3.0s. Determine the velocity of the particle at t = 3.0 s (indicate the direction with the sign of your answer) Determine the acceleration of the particle at t = 30 s (indicate the direction with the sign of your answer) What is the maximum positive coordinate reached by the particle? What is the maximum positive velocity reached by the particle? At what time is it reached? What is the acceleration of the particle is not moving(other than at t = ?(Indicate the direction with the sign of your answer) Determine the average velocity of the particle between t = 0 and t = 3s (Indicate the direction with the sign of your answer)

Explanation / Answer

x=22*t^2-6*t^3

part a:

at t=3 seconds,

x=22*3^2-6*3^3=36 m

part b:
velocity=dx/dt=44*t-18*t^2
at t=3 seconds, velocity=44*3-18*3^2=-30 m/s

part c:

acceleration=dv/dt=44-36*t

at t=3 seconds, acceleration=44-36*3=-64 m/s^2

part d:

maximum positive coordinate will be reached when dx/dt=0 and d^2x/dt^2 <=0

dx/dt=44*t-18*t^2

then dx/dt=0

==>t*(44-18*t)=0

so t=0 or t=2.4444 seconds

at this point d^2x/dt^2=44-36*t=-44 <0

so maximum position is reached at 2.4444 seconds

maximum position=22*2.4444^2-6*2.4444^3=43.8189 m

part e:

at time t=2.4444 seconds, maximum position is reached.

part f:

maximum psoitive velocity will be achieved , when dv/dt=0

==>44-36*t=0

==>t=1.2222 seconds

maximum positive velocity=44*1.2222-18*1.2222^2=26.8889 m/s

part g:

at time t=1.2222 seconds, maximum velocity is reached.

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