A car braked with constant deceleration of 16 ft/sec^2 takes 200 ft to come to a complete stop. How fast was the car traveling when the breaks were first applied?

A) if the brakes were first applied when t = 0 at position s(0) = 0, the initial velocity is an unknown, say V. Solve the differential equation s"(t) = -16 with initial conditions s(0) = 0 and v(0) = V.

B) since the velocity decreases 16 ft/sec^2 every second, it will take t = V/16 seconds for the car to stop. Since it stops after 200 feet, solve s(t) = 200 where t = V/16 for the unknown initial velocity V.

To find the initial velocity of the car when the brakes were first applied, we can use either approach A or B mentioned in the question.

Approach A:
Assume the car's position at time t is given by s(t), and its velocity at time t is given by v(t). We know that the car decelerates at a constant rate of 16 ft/sec^2, which can be expressed as s"(t) = -16 (negative because it is deceleration), where s"(t) represents the second derivative of the position function.

To solve this differential equation, we need two initial conditions: s(0) = 0 (since the car starts at position 0) and v(0) = V (initial unknown velocity). By solving the differential equation s"(t) = -16 with these initial conditions, we can determine the position and velocity functions.

Approach B:
In this approach, we consider the fact that the car's velocity decreases by 16 ft/sec^2 every second. So, after t seconds, the car's velocity will be V - 16t. Since it takes 200 ft to come to a complete stop, the time it takes to stop can be expressed as t = V/16.

Now, we can set up the equation s(t) = 200, where s(t) represents the position function, to find the unknown initial velocity V. Substitute t = V/16 into the equation and solve for V.

Both approaches should yield the same result for the initial velocity V. Once you have obtained the value of V, you can determine how fast the car was traveling when the brakes were first applied.