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Chapter 17: Problem 17

An automobile is traveling at \(25 \mathrm{~m} / \mathrm{s}^{2}\). It takes \(0.3 \mathrm{~s}\) to apply the brakes after which the deceleration is \(6.0 \mathrm{~m} / \mathrm{s}^{2}\). How far does the automobile travel before it stops? a) \(40 \mathrm{~m}\) c) \(50 \mathrm{~m}\) b) \(45 \mathrm{~m}\) d) \(60 \mathrm{~m}\)

Short Answer

Expert verified
Based on the step-by-step solution, determine the total distance an automobile travels after applying the brakes if the initial velocity is 25 m/s, the reaction time is 0.3 seconds, and the deceleration is 6.0 m/s². a) 30 m b) 45 m c) 55 m d) 60 m Answer: d) 60 m

Step by step solution

01

Calculate the distance covered during the reaction time

To find the distance covered during the reaction time, use the formula: \(distance = velocity \times time\). In our case, the velocity is \(25\,\mathrm{m/s}\) and the time is \(0.3\,\mathrm{s}\). So, the distance covered is: \(d_1 = 25\,\mathrm{m/s} \times 0.3\,\mathrm{s} = 7.5\,\mathrm{m}\)
02

Calculate the time it takes to stop the car during the deceleration phase

To determine the time it takes for the car to stop, use the formula: \(final\,\mathrm{velocity} = initial\,\mathrm{velocity} - deceleration \times time\). Since the final velocity is \(0\,\mathrm{m/s}\) and the initial velocity is \(25\,\mathrm{m/s}\), the deceleration is \(6.0\,\mathrm{m/s^2}\). We can rearrange and solve for time: \(time = \frac{initial\,\mathrm{velocity} - final\,\mathrm{velocity}}{deceleration} = \frac{25\,\mathrm{m/s} - 0\,\mathrm{m/s}}{6.0\,\mathrm{m/s^2}} = 4.1667\,\mathrm{s} \)
03

Calculate the distance covered during the deceleration phase

Now, we need to find the distance covered during the deceleration phase. To do this, use the formula: \(distance = initial\,\mathrm{velocity} \times time - \frac{1}{2} \times deceleration \times time^2\). \(d_2 = 25\,\mathrm{m/s} \times 4.1667\,\mathrm{s} - \frac{1}{2} \times 6.0\,\mathrm{m/s^2} \times (4.1667\,\mathrm{s})^2 = 52.083\,\mathrm{m}\)
04

Add the distances covered during the reaction and deceleration phases

To find the total distance traveled, we simply add the two distances calculated in Steps 1 and 3: \(total\,\mathrm{distance} = d_1 + d_2 = 7.5\,\mathrm{m} + 52.083\,\mathrm{m} = 59.583\,\mathrm{m}\)
05

Compare the calculated distance with the answer choices

Finally, compare this calculated distance to the answer choices in the exercise. The closest option to our result of \(59.583\,\mathrm{m}\) is option d) \(60\,\mathrm{m}\). So, the correct answer is d) \(60\,\mathrm{m}\).

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