Question

suppose the position of a particle, s(t) is given by s(t)=t²−8t+30 feet, where t is time in seconds. (a) find the velocity function at time t. s(t)=v(t)= (b) what is the velocity after 3 seconds? feet/second (c) find all value(s) of t (with correct units) for which the particle is at rest (i.e., where v(t)=0). if there are no such values, enter none. if there is more than one value, list them separated by commas. t= seconds video example: solving a similar problem

Explanation:

Part (a)

Step 1: Recall the derivative of position function

The velocity function \( v(t) \) is the derivative of the position function \( s(t) \). For a function \( s(t)=t^2 - 8t + 30 \), we use the power rule for differentiation. The power rule states that if \( y = x^n \), then \( y'=nx^{n - 1} \), and the derivative of a constant is 0.

Step 2: Differentiate \( s(t) \)

Differentiate each term of \( s(t)=t^2-8t + 30 \) with respect to \( t \):

• The derivative of \( t^2 \) is \( 2t \) (using power rule with \( n = 2 \), so \( 2t^{2-1}=2t \)).
• The derivative of \( - 8t \) is \( - 8 \) (using power rule with \( n = 1 \), so \( -8\times1\times t^{1 - 1}=-8 \)).
• The derivative of the constant \( 30 \) is \( 0 \).

So, \( s'(t)=v(t)=2t - 8 \).

Step 1: Substitute \( t = 3 \) into velocity function

We have the velocity function \( v(t)=2t - 8 \). To find the velocity at \( t = 3 \) seconds, we substitute \( t = 3 \) into \( v(t) \).

Step 2: Calculate \( v(3) \)

\( v(3)=2\times3-8=6 - 8=-2 \). The units of velocity are feet per second.

Step 1: Set velocity function to zero

The particle is at rest when \( v(t) = 0 \). We have \( v(t)=2t - 8 \), so we set \( 2t-8 = 0 \).

Step 2: Solve for \( t \)

Add 8 to both sides of the equation: \( 2t=8 \). Then divide both sides by 2: \( t = \frac{8}{2}=4 \).

Answer:

\( v(t)=2t - 8 \)

Part (b)