A 1,200-kg automobile moving at 29 m/s has the brakes applied with a deceleration of 3.0 m/s².How far does the car travel before it stops?

Question

Quizplus · Accepted Answer

The correct answer is A because the formula for the distance traveled by an object with constant acceleration is:d = vi*t + 1/2*a*t^2where:d is the distance traveledvi is the initial velocityt is the timea is the accelerationIn this case, the initial velocity is 29 m/s, the acceleration is -3.0 m/s^2, and the final velocity is 0 m/s. Plugging these values into the formula, we get:d = 29 m/s * t + 1/2*(-3.0 m/s^2)*t^2d = 29t - 1.5t^2To find the distance traveled before the car stops, we need to set the final velocity to 0 m/s and solve for t:0 m/s = 29t - 1.5t^21.5t^2 = 29tt = 19.33 sNow that we know the time it takes for the car to stop, we can plug this value back into the formula for distance traveled to find the distance traveled before the car stops:d = 29 m/s * 19.33 s + 1/2*(-3.0 m/s^2)*(19.33 s)^2d = 560.67 m + 174.67 md = 735.34 mTherefore, the car travels 735.34 m before it stops.

A 1,200-Kg Automobile Moving at 29 M/s Has the Brakes