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gregori [183]
3 years ago
15

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

5 meters per hour. Find the time interval during which the velocity of particle Q is at least 60 meters per hour. Find the distance traveled by particle Q during the interval when the velocity of particle Q is at least 60 meters per hour.

Mathematics
1 answer:
Vladimir79 [104]3 years ago
4 0

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

= \int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt

= By using the Scientific calculator notation;

distance = 106.109 m

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Answer:

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The IQR of the given data is 6. Therefore the correct option is B.

Step-by-step explanation:

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4 0
4 years ago
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Answer:

C

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Two negatives cancel out and turn into a positive.

In the expression 11 - ( -3 5/8 ) there are two negative signs. If the parenthesis are removed, the two negative signs cancel out and turn into a positive and we are left with 11 +3 5/8

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