Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. what is the distance covered before the car comes to a stop? (round your answer to one decimal place.)

Answer 1

The distance covered by the car before it comes to a stop is approximately 303.1 feet.

Step-by-step explanation:

To find the distance covered by the car before it comes to a stop, we can use the equations of motion. The initial velocity of the car is 50 mi/h, which can be converted to feet per second by multiplying by 1.47 ft/s/mi. The deceleration is given as 22 ft/s².

Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered, we can solve for s.

Plugging in the values, the equation becomes 0 = (50*1.47)² + 2*(22)*s. Solving for s, we find that the distance covered before the car comes to a stop is approximately 303.1 feet.

Answer 2

Answer: p[0] = 0 v[0] = 50 mi/h = 50 * 5280 / 3600 ft/s = 50 * (22/15) ft/s = 10 * 22/3 ft/s = 220/3 ft/s a = -50 ft/s^2 a = -50 Integrate to find velocity v = -50t + C 220/3 = -50 * 0 + C 220/3 = C v = 220/3 - 50t Find when v = 0 0 = 220/3 - 50 * t 50 * t = 220/3 t = 220 / 150 t = 22/15 Integrate the velocity function to find the position function p = (220/3) * t - 25 * t^2 + C 0 = (220/3) * 0 - 25 * 0^2 + C 0 = C p = (220/3) * t - 25 * t^2 t = 22/15 p = (220/3) * (22/15) - 25 * (484/225) p = (4840 / 45) - 484 / 9 p = 484 * (10/45 - 1/9) p = 484 * (2/9 - 1/9) p = 484 * (1/9) p = 484/9 p = 53.777777777777777777777777777778 53.8 feet