Evaluate the sine, cosine, and tangent of the angle without using a calculator. θ = − 10π 3

1 answer:

4 0

Determine the quadrant adding or subtracting periods -10pi/3+2pi=-4pi/3+2pi=2pi/3
which is the second quadrant because
first quadrant is between 0 and pi/2
second quadrant is between pi/2 and pi
third quadrant is between pi and 3pi/2
fourth is between 3pi/2 and 2pi
now we find the terminal angle since it is the second quadrant we use theta'=pi-theta=pi-2pi/3=pi/3
sin(pi/3)=sqrt(3)/2 sin is positive in the second quadrant
cos(pi/3)=-0.5 cos is negative in the second quadrant
tan(pi/3)=sin(pi/3)/cos(pi/3)=-sqrt(3) tan is negative in the second quadrant

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A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

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 :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$