Question :
r(t) = 3sint i + 2 cost j - sin2t k at t=π/2
Find Scalar Tangential component of acceleration.
Answer:
Given, r(t) = 3sint i + 2 cost j - sin2t k
Velocity,V = r'(t) = 3cost i - 2sint j - 2cos2t k
V(t=π/2) = -2j + 2k
Acceleration, a = r''(t) = -3sint i - 2cost j + 4sin2t k
a (t=π/2) = -3i
The equation of Scalar Tangential component of acceleration, aT is
aT =( v . a ) / | v |
= (-2j + 2k) . ( -3i ) / root (8)
= 0
Here I am getting (-2j + 2k) . ( -3i ) =0 Is this correct?
Therefore the Scalar Tangential component of acceleration, aT is 0
Is there any mistake in my calculation? I don't have the real answer to this question? Is my answer correct?
No.
Yes.