Question

An object′s position in the plane is defined by vector s of t equals ln of quantity t squared minus 8 times t end quantity comma t cubed over 3 minus 5 over 2 times t squared plus 4 times t period When is the object at rest?

Answer 1

To find when the object is at rest, we need to find when its velocity is zero. The velocity vector v is the derivative of the position vector s with respect to time t:

v(t) = ds/dt

We first find the derivative of each component of s:

ds/dt = (2t - 8) / (t^2 - 8t) i + t^2 / (3 - (5/2)t^2 + 4t) j

To find when the object is at rest, we need to find when the magnitude of the velocity vector is zero:

|v(t)| = sqrt[(ds/dt)^2] = sqrt[(2t-8)^2 / (t^2-8t)^2 + t^4 / (3-(5/2)t^2+4t)^2] = 0

Squaring both sides, we get:

(2t-8)^2 / (t^2-8t)^2 + t^4 / (3-(5/2)t^2+4t)^2 = 0

Since the left-hand side is a sum of squares, it must be non-negative, so the only way for it to be zero is if each term is zero. Thus, we have two equations to solve:

(2t - 8) / (t^2 - 8t) = 0

t^2 / (3 - (5/2)t^2 + 4t) = 0

The first equation implies that 2t - 8 = 0, or t = 4. Substituting t = 4 into the second equation, we get:

(4)^2 / (3 - (5/2)(4)^2 + 4(4)) = 0

This simplifies to:

16 / 3 = 0

which is not true. Therefore, there is no other time when the object is at rest, other than when t = 4.

Thus, the object is at rest when t = 4.