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
Explanation:
Given,
x=4t
2
+5t+16
y=5t
velocity along x-axis
v
x
=
dt
dx
=(8t+5)m/s
acceleration along x-axis,
a
x
=
dt
v
x
=8m/s
velocity alon y-axis
v
y
=
dt
dy
=5m/s
acceleration along y- axis,
a
y
=
dt
v
y
=0m/s
The acceleration of the particle is given by
a
=a
x
i
^
+a
y
j
^
∣
a
∣=
a
x
2
+a
y
2
a=
8
2
+0
2
=8m/s
2
The correct option is A.
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
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