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. Write a function that gives the object's acceleration a(t) in miles per hour per hour. a(t) = ______

Identify Velocity Function: Identify the velocity function and understand that acceleration is the derivative of velocity. Velocity function, v(t) = 4t^3 - 12t^2 + 5t + 10. To find acceleration, we need to differentiate v(t) with respect to t.Differentiate Velocity Function: Differentiate each term of v(t) separately. Differentiate 4t^3: The derivative is 12t^2. Differentiate -12t^2: The derivative is -24t. Differentiate 5t: The derivative is 5. 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.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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) = ______

View step-by-step help

Home

Math Problems

Calculus

Relate position, velocity, speed, and acceleration using derivatives

Full solution

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: $a(t) = 12t^2 - 24t + 5$ .