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3 years ago

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The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t2]i^

− (0.0150m/s3)t3j^.
At what value of t does the velocity vector of the insect make an angle of 36.0 ∘ clockwise from the x-axis? ... t=??

2 answers:

Arlecino [84]3 years ago

7 0

The solution for this problem is:

r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters

v = dr/dt = [0.180t i - 0.0450t² j] m/s²

tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect

White raven [17]3 years ago

7 0

The velocity vector of the insect make an angle of 36.0° clockwise from the x-axis at t = 2.91 s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

$r = [2.90 +(0.0900t^2)]i - [(0.0150)t^3]j$

To find the velocity function, we will derive the position function above.

$v = dr/dt = [0 +(0.0900)(2t)]i - [(0.0150)(3t^2)]j$

$v = [0.18t]i - [0.045t^2]j$

Next to calculate the angle , we use this following formula:

$\tan \theta = \frac{v_y}{v_x}$

$\tan 36^o = \frac{0.045t^2}{0.18t}$

$\tan 36^o = \frac{0.045t}{0.18}$

$t = \frac{\tan 36^o \times 0.18}{0.045}$

$t \approx 2.91 ~ s$

<h2>Conclusion:</h2>

The velocity vector of the insect make an angle of 36.0° clockwise from the x-axis at t = 2.91 s

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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