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An object travels along a straight line. The function $v(t) = 4\sqrt{t} - 3$ gives the object's velocity, in kilometers per hour, at time $t > 0$ hours. $\newline$ Write a function that gives the object's acceleration $a(t)$ in kilometers per hour per hour. $\newline$ a(t) = ______

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Relate position, velocity, speed, and acceleration using derivatives

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Q. An object travels along a straight line. The function

v(t) = 4\sqrt{t} - 3

gives the object's velocity, in kilometers per hour, at time

t > 0

hours.

\newline

Write a function that gives the object's acceleration

a(t)

in kilometers per hour per hour.

\newline

a(t) = ______

Differentiate Velocity Function: To find the acceleration function $a(t)$ , we need to differentiate the velocity function $v(t)$ with respect to time $t$ . The velocity function given is $v(t) = 4\sqrt{t} - 3$ .
Apply Power Rule: Differentiate $v(t) = 4\sqrt{t} - 3$ . Using the power rule, $\sqrt{t}$ can be rewritten as $t^{(1/2)}$ . So, the derivative of $t^{(1/2)}$ is $(1/2)t^{(-1/2)}$ . Applying the constant multiple rule, the derivative of $4t^{(1/2)}$ is $4 \times (1/2)t^{(-1/2)} = 2t^{(-1/2)}$ . The derivative of a constant $(-3)$ is $0$ .
Acceleration Function Calculation: Therefore, the acceleration function $a(t) = 2t^{(-1/2)}$ , which simplifies to $\frac{2}{\sqrt{t}}$ .

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