Answer:
The time interval when [tex]V_Q(t) \geq 60[/tex] is at [tex]1.866 \leq t \leq 3.519[/tex]
The distance is 106.109 m
Step-by-step explanation:
The velocity of the second particle Q moving along the x-axis is :
[tex]V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)[/tex]
So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.
We are also to that :
[tex]V_Q(t) \geq 60[/tex] between [tex]0 \leq t \leq 4[/tex]
The schematic free body graphical representation of the above illustration was attached in the file below and the point when [tex]V_Q(t) \geq 60[/tex] is at 4 is obtained in the parabolic curve.
So, [tex]V_Q(t) \geq 60[/tex] is at [tex]1.866 \leq t \leq 3.519[/tex]
Taking the integral of the time interval in order to determine the distance; we have:
distance = [tex]\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt[/tex]
= [tex]\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt[/tex]
= By using the Scientific calculator notation;
distance = 106.109 m
