A car is traveling \( 88 \mathrm{ft} / \mathrm{sec} \) at the moment the brakes are applied. The deceleration is a constant \( -10 \mathrm{ft} / \mathrm{sec}^{2} \). The brakes were applied at \( t=0 \) and the initial position is 0 ft . What is the position at \( t=5 \) ? 315 ft . How much time will it take the car to stop? \( \frac{[?]}{[\square]} \) seconds

To find out how much time it takes for the car to stop, we can use the formula for acceleration. The formula is: \[ v = v_0 + at \] Where: - \( v \) is the final velocity (0 ft/sec when the car stops), - \( v_0 \) is the initial velocity (88 ft/sec), - \( a \) is the acceleration (-10 ft/sec²), - \( t \) is the time in seconds. Setting up the equation: \[ 0 = 88 - 10t \] Rearranging it gives: \[ 10t = 88 \implies t = \frac{88}{10} = 8.8 \text{ seconds}. \] So, the car will take **8.8 seconds** to stop! Did you know that the concept of braking distance and time is not just for cars? It's also crucial for airplanes during landing. Pilots must account for similar concepts as they come in for a landing, adjusting for wind speed, landing angle, and runway conditions to ensure a safe stop. If you’re ever interested in the math behind everyday scenarios, check out online resources related to kinematics! Websites like Khan Academy offer lessons on motion equations, making it fun and engaging to grasp the fundamentals of motion, speed, and acceleration without feeling like you're sitting in a typical school lecture!