A car initially going $50 \mathrm{ft} / \mathrm{sec}$ brakes at a constant rate (constant negative acceleration, also called deceleration), and the car comes to a stop in 5 seconds.
a. Graph the velocity from $t=0$ to $t=5$.
b. How far does the car travel?
c. How far does the car travel if its initial velocity is doubled, but the car decelerates at the same constant rate?

Step 1 ：The problem gives us that a car initially going at a speed of 50 ft/sec brakes at a constant rate and comes to a stop in 5 seconds.

Step 2 ：This means that the car's velocity decreases linearly from 50 ft/sec to 0 ft/sec over 5 seconds. Therefore, the deceleration rate is \(-10 \, \text{ft/sec}^2\).

Step 3 ：We can calculate the distance the car travels by integrating the velocity function over the time interval [0, 5].

Step 4 ：Let's denote time as \(t\), velocity as \(v\), and distance as \(d\). We have \(v = 50 - 10t\).

Step 5 ：By integrating the velocity function over the time interval [0, 5], we find that \(d = 125\).

Step 6 ：So, the car travels \(\boxed{125}\) feet.