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3 3 The braking distance of a car is directly proportional to its initial kinetic energy. The braking distance of a car is 18m when its initial speed is 10ms^-1 What is the braking distance of the car under the same conditions when its initial speed is 25ms^-1

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3
3 The braking distance of a car is directly proportional to its initial kinetic energy.
The braking distance of a car is 18m when its initial speed is 10ms^-1
What is the braking distance of the car under the same conditions when its initial speed is 25ms^-1

3 3 The braking distance of a car is directly proportional to its initial kinetic energy. The braking distance of a car is 18m when its initial speed is 10ms^-1 What is the braking distance of the car under the same conditions when its initial speed is 25ms^-1

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SaffronVeteran · Tutor for 10 years

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<p> Under the same conditions, a car travelling initially at 25 ms⁻¹ will take 112.5 meters to stop.</p>

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<p> This question necessitates the understanding of the role kinetic energy plays in an object's motion, primarily when involved in braking or stopping. As stated in the question, the braking distance of a car is directly proportional to the car's kinetic energy, subject to the same braking conditions.<br /><br />The kinetic energy depends upon the mass and the square of the magnitude of velocity. The formula for kinetic energy is:<br /><br />K.E = 0.5m(v^2)<br /><br />Although a value for the car's mass isn't directly given, we seem to work through this problem via analysing the question in proportional terms, deeming mass as not necessary for this particular question.<br /><br />Given that the initial speed and the corresponding braking distance of a vehicle is set, we can find the ratio of new braking distance by square of speed ratios.<br /><br />We already know that from the given data: when speed υ_1 = 10 m/s, distance d_1 = 18 m; we ought to solve for d_2 where speed υ_2= 25 m/s.<br /><br />The subsequent equation encapsulates appropriate working assumptions concerning equality, glyphs must in a manner retain an equivalent relationship:<br /><br />squares of speed ratios υ_1 ² : υ_2 ² equal to ratio of distances d_1 : d_2<br /><br />Subbing in the velocites as well as equivalent distance, comparatively extracting a ratio for the basis second distance can be attained. From here rearrange the equation to solve for d_2.</p>
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