Question

Suppose a particle moves along a curve r= (t³-4t) i + (t²+4t) j + (8t²-3t³) k. What are the magnitudes acceleration along the tangential and normal components of its acceleration when t=2?​

Answer 1

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|