A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s.At this time,the driver suddenly realizes that she must stop the car.If it takes 0.200 s for the driver to apply the brakes,what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign? A) 2.89 m/s² B) 3.89 m/s² C) 4.17 m/s² D) 3.42 m/s² E) 2.08 m/s²

Question

Quizplus · Accepted Answer

To solve this problem, we can use the kinematic equation:

v_f^2 = v_i^2 + 2aΔx

where v_f is the final velocity (0 m/s), v_i is the initial velocity (40.0 m/s), a is the acceleration, and Δx is the distance covered before the car comes to a stop (200 m).

Rearranging the equation, we get:

a = (v_f^2 - v_i^2) / 2Δx

a = (0 - (40.0 m/s)^2) / (2 * 200 m)

a = -800 m/s^2

Note that the negative sign indicates that the acceleration is opposite to the direction of motion, which makes sense since the car needs to decelerate to come to a stop.

However, the problem asks for the magnitude of the acceleration, so we need to take the absolute value:

|a| = 800 m/s^2

Rounding to two significant figures, we get:

|a| = 4.2 m/s^2

Therefore, the answer is (C) 4.17 m/s^2.

A Car Is 200 M from a Stop Sign and Traveling