A particle starts at the origin(xi=0) at time t=0. Once it begins moving, the po

Question

A particle starts at the origin(xi=0) at time t=0. Once it begins moving, the position of the particle as a function of time is given by: x(t)=(2.2 m/s)t+(-2.6 m/s^2)t^2.

a) Find the particles velocity as a function of time , v(t).

v(t)=dx/dt=(2.2 m/s)+(-6.7 m/s^2)t

b) Find the time that the particle reverses direction.

c) Find the position of the particle where the particle reverses direction.

d) Find the amount of time that passes until the particle returns to the origin.

e) Find the velocity of the particle once it returns to the origin.

Explanation / Answer

x(t) = 2.2t-2.6t^2
a) v(t) = dx/dt = 2.2 m/s + (-5.2 m/s^2)t
b) at the time when particle reverses its motion, velocity will be zero, so
2.2-5.2t = 0 => t = 0.4 sec
c) Position at this time, x(0.4) = 2.2*0.4-2.6*0.4^2 = 0.465 m
d) when it returns back to origin, x(t) = 0
2.2t-2.6t^2 = 0
t = 0 or t = 0.846 sec
So the time spent = 0.846 sec
e) velocity at this instant= 2.2-5.2*0.846 = -2.2 m/s.

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