A mosquito is moving with a velocity v = 0.5t² i + 3t j + 9 k m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2s?

Option 1 - <p>tan⁻¹(2/3) from x-axis</p>
Option 2 - <p>tan⁻¹(5/2) from y-axis<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>tan⁻¹(5/2) from x-axis</p>
Option 4 - <p>tan⁻¹(2/3) from y-axis</p>
26 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 4
Detailed Solution:

v = 0.5t² i + 3t j + 9 k m/s ⇒ a = dv/dt = (ti + 3j) m/s²
At t=2sec,  v = 2i + 6j + 9k m/s and a = (2i + 3j) m/s²
The direction of acceleration of mosquito after 2s is given by the angle θ with the y-axis, where tanθ = a? /a? = 2/3.
So, the direction is tan? ¹ (2/3) from the y-axis.

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Physics Motion in Plane 2025

Physics Motion in Plane 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering