Deceleration A car slows down with an acceleration of a(t) = −15 ft/s². Assume v(0)=60 ft/s,s(0)=0, and t is measured in seconds.


b. How far does the car travel in the time it takes to come to rest?

Verified step by step guidance

1. Identify the given information: acceleration \(a(t) = -15\) ft/s² (constant), initial velocity \(v(0) = 60\) ft/s, and initial position \(s(0) = 0\) ft.
2. Find the time \(t\) when the car comes to rest by setting the velocity \(v(t) = 0\). Use the relationship between acceleration and velocity: \(v(t) = v(0) + \int_0^t a(\tau) \, d\tau\).
3. Since acceleration is constant, express velocity as \(v(t) = v(0) + a \cdot t = 60 - 15t\). Solve for \(t\) when \(v(t) = 0\).
4. Find the position function \(s(t)\) by integrating the velocity function: \(s(t) = s(0) + \int_0^t v(\tau) \, d\tau\). Substitute \(v(\tau) = 60 - 15\tau\) and \(s(0) = 0\).
5. Evaluate \(s(t)\) at the time found in step 3 to determine the total distance traveled until the car comes to rest.

Key Concepts

Kinematic Equations and Integration

Kinematic equations relate acceleration, velocity, and displacement over time. Since acceleration is given as a function of time, integrating acceleration yields velocity, and integrating velocity yields displacement. This process helps find how far the car travels before stopping.

Initial Conditions in Motion Problems

Initial conditions like initial velocity v(0) and initial position s(0) are essential to solve differential equations uniquely. They allow determination of constants of integration, ensuring the solution matches the physical scenario of the car starting at 60 ft/s and position zero.

Time to Come to Rest

The time when the car comes to rest is when velocity equals zero. Using the velocity function derived from acceleration, setting v(t) = 0 and solving for t gives the stopping time. This time is then used to find the total distance traveled.