Q1.a particle moves in x-y plane according to the law x=4t and y =t(8-t) where x,y are in meters and 't' is in second . calculate the time in second after which velocity and acceleration becomes mutually perpendicular .
Answers
Step-by-step explanation:
SOLUTION
GIVEN
A particle moves in x-y plane according to the law x = 4t and y = t( 8 - t) where x,y are in meters and 't' is in second .
TO DETERMINE
The time in second after which velocity and acceleration becomes mutually perpendicular .
EVALUATION
Here it is given that a particle moves in x-y plane according to the law x = 4t and y = t( 8 - t) where x,y are in meters and 't' is in second
Differentiating we get -
Component of velocity along x axis = 4 m/s
Component of velocity along y axis = 8 - 2t m/s
Differentiating we get -
Component of acceleration along x axis = 0
Component of acceleration along y axis
= - 2 m/s²
Now it is given that velocity and acceleration are mutually perpendicular
So we get
( 0 × 4 ) + - 2 × ( 8 - 2t ) = 0
⇒ - 2 × ( 8 - 2t ) = 0
⇒ ( 8 - 2t ) = 0
⇒ 2t = 8
⇒ t = 4
FINAL ANSWER
After 4 seconds velocity and acceleration becomes mutually perpendicular
━━━━━━━━━━━━━━━━
Learn more from Brainly :-
1.1. If two vectors are perpendicular to each other then which of these is zero
(a) scalar product or dot product (b) vector...
https://brainly.in/question/31020619
2. two vectors of magnitude 4 u and 3 u are such that their scalar product is zero. Find their resultant.
https://brainly.in/question/30302455