Question

A car starting from rest moves in a straight line with a constant acceleration of 2.20 m/s² for 14.0 s, then slows down to a stop with a deceleration of 2.90 m/s². How far does it travel?

Answer 1

The car travels a total distance of 22.480 m.

Step-by-step explanation:

To find the distance traveled by the car, we need to calculate the distance during the constant acceleration phase and the deceleration phase separately.

During the constant acceleration phase, we can use the equation:

d = v0t + 1/2at2

Substituting the given values:

d = 0 * 14 + 1/2 * 2.20 * (14)2

d = 20.580 m

During the deceleration phase, the final velocity is 0 m/s and the acceleration is -2.90 m/s²:

vf2 = v02 + 2ad

Substituting the relevant values:

0 = (2.20)2 + 2 * (-2.90) * d

d = 1.900 m

The total distance traveled by the car is the sum of the distances during the two phases:

Total Distance = 20.580 + 1.900 = 22.480 m