How Far Apart Are the Stop Signs?

[kingpin29]

Homework Statement

A car starts from rest at a stop sign. It accelerates at 4.0 m/s^2 for 6.0 s, coasts for 1.9 s, and then slows down at a rate of 3.0 m/s^2 for the next stop sign.
How far apart are the stop signs?

Homework Equations

The Attempt at a Solution

 

[kuruman]

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[tomcue]

As a scientist, it is important to approach this question with a systematic and analytical mindset. First, we can use the equations of motion to calculate the distance traveled during each phase of the car's motion.

During the acceleration phase, we can use the equation d = (1/2)at^2 to calculate the distance traveled, where d is the distance, a is the acceleration, and t is the time. Plugging in the given values, we get d = (1/2)(4.0 m/s^2)(6.0 s)^2 = 72.0 m.

During the coasting phase, the car maintains a constant velocity, so the distance traveled can be calculated using the equation d = vt, where v is the velocity and t is the time. Since the car is coasting, the velocity is constant at 4.0 m/s, and the time is 1.9 s, we get d = (4.0 m/s)(1.9 s) = 7.6 m.

For the deceleration phase, we can use the equation v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity (which is the same as the coasting velocity of 4.0 m/s), a is the acceleration (-3.0 m/s^2), and d is the distance. Solving for d, we get d = (v^2 - u^2)/(2a) = (0 - 4.0 m/s)^2/(2(-3.0 m/s^2)) = 2.67 m.

Therefore, the total distance traveled between the two stop signs is 72.0 m + 7.6 m + 2.67 m = 82.27 m. However, it is important to note that this is an ideal scenario and does not take into account factors such as friction, air resistance, and other external forces that may affect the car's motion. In reality, the distance between the two stop signs may vary slightly.