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Chapter 3: Problem 21

A car moving initially at a speed of \(50 \mathrm{mi} / \mathrm{h}(\approx 80 \mathrm{~km} / \mathrm{h})\) and weighing \(3000 \mathrm{lb}(\approx 13,000 \mathrm{~N})\) is brought to a stop in a distance of \(200 \mathrm{ft}(\approx 61 \mathrm{~m})\). Find \((a)\) the braking force and \((b)\) the time required to stop. Assuming the same braking force, find \((c)\) the distance and \((d)\) the time required to stop if the car were going \(25 \mathrm{mi} / \mathrm{h}(\approx 40 \mathrm{~km} / \mathrm{h})\) initially.

Short Answer

Expert verified
The braking force is approximately 11218.79 Newtons (but considered negative if the motion's direction is set as positive). The time required to bring the car to stop from 50 mi/h is about 2.71 seconds. If the car's initial speed were 25 mi/h instead, it would require about 7.55 meters to stop and would do so in approximately 1.35 seconds.

Step by step solution

01

Convert units to the SI system

The speed should be converted from miles per hour to meters per second, the weight from pounds to Newtons, and the distance from feet to meters. Using unit conversion: speed \(v1 = 50 \ mi/h = 22.35 \ m/s\), weight \( W = 3000 \ lb = 13344.5 \ N\), stopping distance \(d1 = 200 \ ft = 60.96 \ m\). The gravitational force is \( g = 9.8 \ m/s^2\). The mass of the car \(m = W / g = 1360.46 \ kg\).
02

Find the braking force

Braking force can be obtained from the equation of motion: \(v^2 = u^2 - 2a x\), where \(a\) is the deceleration, \(u\) is the initial velocity, \(v\) is the final velocity, and \(x\) is the displacement(or distance travelled). Solve this formula for \(a = (u^2 - v^2) / (2 * x)\). The result is \(a = -22.35^2 / (2 * 60.96) = -8.25 \ m/s^2\). The negative sign represents braking or slowing down. The force required for this deceleration is \(f = m * a = 1360.46 * -8.25 = -11218.79 \ N\). By law of motion, the force of braking is exerted in the opposite direction to the motion, hence the negative sign.
03

Find the time needed to stop

Time can be obtained using the first equation of motion: \(v = u + at\), where \(t\) is the time, rearranging for \(t\) gives \(t = (v - u) / a = (0 - 22.35) / -8.25 = 2.71 \ seconds\).
04

Assuming the same braking force, find the distance and time required to stop if the car were going 25 mi/h initially

Repeat Step 1, but for \(u2 = 25 \ mi/h = 11.176 \ m/s\). Maintaining the same deceleration \(a = -8.25 \ m/s^2\), apply the equation of motion to find the new stopping distance \(x = (u^2 - v^2) / (2a) = (11.176^2 - 0) / (2 * -8.25) = 7.55 \ m\). Then use the first equation of motion to find the new stopping time: \(t = (v - u) / a = (0 - 11.176) / -8.25 = 1.35 \ seconds\).

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