On a dry road, a car with good tires may


On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.92 m/s^2. (a) How long does such a car, initially traveling at 24.6 m/s, take to stop? (b) How far does it travel in this time?

Solution:

To solve this question, we will use the following equation:

v=v_0+at where v is final velocity, v_0 is initial velocity, a is acceleration and t is time.

a)We will now substitute the values given to us in the question. Note that v is equal to 0 because the car comes to a complete stop and the acceleration is negative because it’s deceleration.

0=24.6m/s+(-4.92m/s^2)t

t=\frac{24.6m/s}{-4.92m/s^2}

t=5.00 s

Therefore, the time it took for the car to stop is 5 seconds.

 

b) To find the distance the car travels in this time, we will use the following equation:

v^2=v_0^2+2ax where x is the position or distance.

x=-\frac{(24.6 m/s)^2}{2(-4.92 m/s^2)}

x=61.5 m

Therefore, the total distance the car traveled while the brakes were applied was 61.5 m.

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 28.

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