Question

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=??

Answer 1

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

Answer 2

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

Further explanation

Acceleration is rate of change of velocity.

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

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

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

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

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:

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

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

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

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

$t \approx 2.91 ~ s$

Conclusion:

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

Learn more

• 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

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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