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Chapter 2: Problem 8

A car moving at \(60 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(4.0 \mathrm{~s}\). What was its average deceleration? a) \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) b) \(15 \mathrm{~m} / \mathrm{s}^{2}\) c) \(4.2 \mathrm{~m} / \mathrm{s}^{2}\) d) \(41 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
A) 2.50 m/s^2 B) 3.17 m/s^2 C) 4.17 m/s^2 D) 5.17 m/s^2 Answer: C) 4.17 m/s^2

Step by step solution

01

Convert initial velocity from km/h to m/s

First, we need to convert the initial velocity from kilometers per hour to meters per second. We can do this using the conversion factor 1000 meters per kilometer and 3600 seconds per hour. Multiply the initial velocity by 1000/3600: \(60 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 16.67 \frac{\mathrm{m}}{\mathrm{s}}\)
02

Apply the formula for acceleration

Now that we have the initial velocity in meters per second, we can use the formula for acceleration: \(a = \frac{v_f - v_i}{t}\) where \(a\) is the acceleration, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time it takes for the change in velocity to occur.
03

Substitute the given values and solve for average deceleration

Plug in the given values: \(v_f = 0 \frac{\mathrm{m}}{\mathrm{s}}\), \(v_i = 16.67 \frac{\mathrm{m}}{\mathrm{s}}\), and \(t = 4.0 \mathrm{~s}\): \(a = \frac{0 \mathrm{~m/s} - 16.67 \mathrm{~m/s}}{4.0 \mathrm{~s}}\) \(a = -4.17 \frac{\mathrm{m}}{\mathrm{s}^{2}}\) The average deceleration is \(-4.17 \frac{\mathrm{m}}{\mathrm{s}^{2}}\). Since deceleration is negative acceleration, and we are looking for the magnitude of deceleration, we can drop the negative sign and obtain the answer: \(4.17 \frac{\mathrm{m}}{\mathrm{s}^{2}}\) The correct answer is closest to option C).

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Most popular questions from this chapter

A ball is thrown directly downward, with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\), from a height of \(50.0 \mathrm{~m}\). After what time interval does the ball strike the ground?

The vertical position of a ball suspended by a rubber band is given by the equation $$ y(t)=(3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s}-0.31)-(0.2 \mathrm{~m} / \mathrm{s}) t+5.0 \mathrm{~m} $$ a) What are the equations for velocity and acceleration for this ball? b) For what times between 0 and \(30 \mathrm{~s}\) is the acceleration zero?

An F-14 Tomcat fighter jet is taking off from the deck of the USS Nimitz aircraft carrier with the assistance of a steam-powered catapult. The jet's location along the flight deck is measured at intervals of \(0.20 \mathrm{~s} .\) These measurements are tabulated as follows: $$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0.00 & 0.20 & 0.40 & 0.60 & 0.80 & 1.00 & 1.20 & 1.40 & 1.60 & 1.80 & 2.00 \\ \hline x(\mathrm{~m}) & 0.0 & 0.70 & 3.0 & 6.6 & 11.8 & 18.5 & 26.6 & 36.2 & 47.3 & 59.9 & 73.9 \\ \hline \end{array} $$ Use difference formulas to calculate the jet's average velocity and average acceleration for each time interval. After completing this analysis, can you say if the F- 14 Tomcat accelerated with approximately constant acceleration?

A bank robber in a getaway car approaches an intersection at a speed of 45 mph. Just as he passes the intersection, he realizes that he needed to turn. So he steps on the brakes, comes to a complete stop, and then accelerates driving straight backward. He reaches a speed of \(22.5 \mathrm{mph}\) moving backward. Altogether his deceleration and reacceleration in the opposite direction take \(12.4 \mathrm{~s} .\) What is the average acceleration during this time?

The 2007 world record for the men's 100 -m dash was \(9.77 \mathrm{~s}\). The third-place runner crossed the finish line in \(10.07 \mathrm{~s}\). When the winner crossed the finish line, how far was the third-place runner behind him? a) Compute an answer that assumes that each runner ran at his average speed for the entire race. b) Compute another answer that uses the result of Example 2.3, that a world- class sprinter runs at a speed of \(12 \mathrm{~m} / \mathrm{s}\) after an initial acceleration phase. If both runners in this race reach this speed, how far behind is the third-place runner when the winner finishes?
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