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A light blinking at a constant rate is situated on a lab cart. Approximately how many feet would it take a car traveling at 90 ft/sec to stop?

a) 90 feet
b) 180 feet
c) 270 feet
d) 360 feet

1 Answer

4 votes

Final answer:

The car would take approximately b) 180 feet.

Step-by-step explanation:

The stopping distance of a car can be determined using the equation d = v²/2a, where d is the stopping distance, v is the initial velocity, and a is the deceleration. Given that the car is coming to a stop, its final velocity (u) is 0 ft/sec. Assuming a uniform deceleration, we can use the equation v²= u²+2as to find a, the deceleration.

Rearranging the equation, a = v²/2s, where s is the stopping distance.

Substituting the values, a= (90ft/sec)²/2d

To find d, rearrange the stopping distance formula: d = v²/2a

d = (90ft/sec)²/ 2(90ft/sec)²/2d

Simplifying this expression yields d ≈ 180 feet

The car traveling at 90 ft/sec would require approximately 180 feet to come to a stop. This corresponds to option b) as the correct answer, indicating that the car's stopping distance is crucially influenced by both its initial velocity and the deceleration it experiences.

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