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 will take the car to stop, we can use the formula for deceleration. Given the initial velocity \( u = 88 \mathrm{ft/sec} \) and a deceleration \( a = -10 \mathrm{ft/sec}^2 \), we can use the formula: \[ v = u + at \] where \( v \) is the final velocity (0 ft/sec since the car stops). Rearranging for \( t \) gives: \[ 0 = 88 + (-10)t \implies 10t = 88 \implies t = \frac{88}{10} = 8.8 \text{ seconds} \] The car will take \( \frac{88}{10} \) seconds or 8.8 seconds to come to a complete stop. Now, if you want to impress your friends with some trivia, consider this: the average stopping distance for a car is often calculated as \( \frac{v^2}{2a} \). So in this case, \( \frac{88^2}{2 \times 10} = 384 \text{ feet} \). That's a good bit longer than the distance traveled in 5 seconds! For those intrigued by vehicular dynamics, delving into topics such as momentum, friction, and the physics of braking can really spice things up. Check out “Physics of the Everyday World” for a fun mix of theory and practice, or watch some educational YouTube videos on how different factors like tire conditions affect stopping distances.