Question

a car starting from rest moves with constant acceleration of 2.0 m/s2 for 10 s, then travels with constant speed for another 10 s, and then finally slows to a stop with constant acceleration of -2.0 m/s2. how far does it travel? group of answer choices 500 m 400 m 200 m 300 m

Answer 1

The car travels a total distance of 32 m.

Step-by-step explanation:

The distance traveled by the car can be found by calculating the sum of the distances traveled in each segment of the motion.

In the first 10 seconds, the car moves with constant acceleration and covers a distance equal to the average velocity multiplied by the time: (0 m/s + 2.0 m/s) / 2 * 10 s = 10 m.

In the next 10 seconds, the car travels with constant speed. Therefore, the distance traveled is equal to the speed multiplied by the time: 2.0 m/s * 10 s = 20 m.

Finally, in the last segment, the car slows down with a constant acceleration and comes to a stop after 10 seconds. The distance traveled can be calculated using the equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. Rearranging the equation, we get: 0^2 = (2.0 m/s)^2 + 2 * (-2.0 m/s^2) * s. Solving for s, we find s = 2.0 m.

Adding up the distances traveled in each segment, we find that the car travels a total distance of 10 m + 20 m + 2.0 m = 32 m.