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A car is traveling at 36.6 m/s on Hwy 360 when the driver must come to a stop. If the driver accelerates at a rate of -6.8 m/s², how far will he travel before he stops?

a. 103.4 m
b. 154.5 m
c. 205.7 m
d. 256.8 m

Answer :

Final answer:

Using the kinematic equation for uniformly accelerated motion, the correct stopping distance for a car decelerating at -6.8 m/s² from an initial speed of 36.6 m/s is 98.5 meters, which does not match any of the provided options.

Explanation:

To calculate the distance a car will travel before coming to a stop when it is decelerating, we use the kinematic equation for uniformly accelerated motion, which is:

d = v^2 / (2a)

Here, d represents the distance, v is the initial velocity, and a is the acceleration (deceleration, in this case, so it will be a negative value).

According to the problem:

  • Initial velocity v = 36.6 m/s
  • Deceleration a = -6.8 m/s²

Plugging these values into the equation gives us:

d = (36.6 m/s)^2 / (2 × -6.8 m/s²)

d = 1339.56 m^2/s^2 / (-13.6 m/s²)

d = -98.5 m

Since distance cannot be negative in this context, we take the absolute value which gives us:

d = 98.5 m

None of the provided options match the calculated result. Therefore, the correct stopping distance, using the given initial speed and deceleration rate, is 98.5 meters.

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Rewritten by : Jeany

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