If the car's deceleration has a magnitude of 6 m/s^2, how far does it travel while reducing speed?

Question

The driver of a car traveling at 30 m/s needs to quickly reduce the car's speed to 20 m/s to round a curve on the highway. If the car's deceleration has a magnitude of 6 m/s^2, how far does it travel while reducing speed?

Answer 1

To decrease velocity 10 m/s it would take 1.66 seconds of deceleration. Vfinal = accel * time + Voriginal

20 = -6*t + 30

-10 = -6*t

10/6 = t

Since the deceleration is constant, the distance traveled during deceleration is the average speed, (30+20)/2 = 25 m/s, multiplied by the number of seconds it takes to stop.

So distance traveled = 1.666 * 25 = 41.5 meters during deceleration.

If you wanted to do it the hard way, you could also get the same answer by using the equation for position at time = 1.667 seconds. x = 1/2 a*t² + V*t + xoriginal

x = (0.5) * -6 * 1.667² + 30 * 1.667 = 41.5