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An object travels along a straight line. The function $v(t) = 4t^3 - 12t^2 + 5t + 10$ gives the object's velocity, in miles per hour, at time $t$ hours. $\newline$ Write a function that gives the object's acceleration $a(t)$ in miles 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) = 4t^3 - 12t^2 + 5t + 10

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

t

hours.

\newline

Write a function that gives the object's acceleration

a(t)

in miles per hour per hour.

\newline

a(t) = ______

Identify Velocity Function: Identify the velocity function and understand that acceleration is the derivative of velocity. $\newline$ Velocity function, $v(t) = 4t^3 - 12t^2 + 5t + 10$ . $\newline$ To find acceleration, we need to differentiate $v(t)$ with respect to $t$ .
Differentiate Velocity Function: Differentiate each term of $v(t)$ separately. $\newline$ Differentiate $4t^3$ : The derivative is $12t^2$ . $\newline$ Differentiate $-12t^2$ : The derivative is $-24t$ . $\newline$ Differentiate $5t$ : The derivative is $5$ . $\newline$ Differentiate $10$ : The derivative is $0$ , since the derivative of a constant is zero.
Combine Differentiated Terms: Combine all the differentiated terms to form the acceleration function $a(t)$ . $a(t) = 12t^2 - 24t + 5 + 0$ Simplify to: $a(t) = 12t^2 - 24t + 5$ .

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