A particle moves along the curve r= (t⁵-4t)i+(8t²-t³)k
Answers
Step-by-step explanation:
Answer
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RUPAMCHAKRABORTY
Answer:
Again differentiating (1)we get acceleration
=(6t)i+2j+(16–18t)k
at t=2
a=6×2i+2j+(16–18×2)k
=12i+2j-20k
Now
Tangential component of acceleration is that component which is along the direction of velocity and perpendicular or radial component is that which is perpendicular to the velocity.
Now for calculating tangential component we take the help of diagram
Let us assume that angle between acceleration and velocity at t=2 is theeta
as you can see acos@ is tangential component of acceleration which is along the direction of velocity and asin@ is perpendicular component of acceleration.
A. V(dot product) =|A||V|cos@
|A|cos@=A. V÷|V|
A. V=(12i+2j-20k).(8i+8j-4k)
=12×8+2×8+(-20×-4)
=96+16+80
A. V=192
acos@=A. V÷|V|