Question

an object moves along a straight line so that at any time t, for 0 ≤ t ≤ 8, its position is given by x(t)=5 + 4t - t². for what value of t is the object at rest? a t = 2 b t = 4 c t = ½ d t = 5

Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(x(t)\).
Given \(x(t)=5 + 4t-t^{2}\), using the power rule \(\frac{d}{dt}(t^{n})=nt^{n - 1}\) and \(\frac{d}{dt}(c)=0\) (where \(c\) is a constant), we have \(v(t)=x^{\prime}(t)=4-2t\).

Step2: Set the velocity equal to zero

An object is at rest when \(v(t) = 0\).
Set \(4-2t=0\).
Solve for \(t\):

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Answer:

A. \(t = 2\)