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Chapter 5: Problem 15

For same braking force the stopping distance of a vehicle increases from \(15 \mathrm{~m}\) to \(60 \mathrm{~m}\). By what factor the velocity of vehicle has been changed (A) 2 (B) 3 (C) 4 (D) \(3 \sqrt{5}\)

Short Answer

Expert verified
The factor by which the velocity of the vehicle has changed is 2. The correct answer is (A) 2.

Step by step solution

01

Set up the stopping distance formula for both distances

Given the stopping distance formula: \( d = \frac{v^2}{2a} \) For the first stopping distance (15m), the formula will be: \( d_1 = \frac{v_1^2}{2a} \) For the second stopping distance (60m), the formula will be: \( d_2 = \frac{v_2^2}{2a} \)
02

Set up the proportion to determine the factor by which the velocity changes

Since we're given that the stopping distances increase (from 15m to 60m) but the braking force (and acceleration) remains the same, we can write the following proportion: \( \frac{v_1^2}{2a} = 15 \) and \( \frac{v_2^2}{2a} = 60 \) Now, we're going to set up a proportion to compare the two velocities: \( \frac{v_1}{v_2} = \sqrt{\frac{d_1}{d_2}} \)
03

Solve for the factor by which the velocity changes

Now we plug in the given values for the stopping distances (15m and 60m): \( \frac{v_1}{v_2} = \sqrt{\frac{15}{60}} \) Simplify the fraction under the square root: \( \frac{v_1}{v_2} = \sqrt{\frac{1}{4}} \) Find the square root of 1/4: \( \frac{v_1}{v_2} = \frac{1}{2} \) This means that 2 times the initial velocity equals the final velocity: \( 2v_1 = v_2 \) So, the factor by which the velocity of the vehicle has changed is 2. The correct answer is (A) 2.

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