Acar traveling at a speed of v can brake to an emergency stop in a distance x. assuming all other driving conditions are all similar, if the traveling speed of the car doubles, the stopping distance will be

We will use the formula: 

2as = v² - u²; where the final velocity v is 0, the distance s will be represented by x and the initial velocity will be represented by v. So:

2ax = -v²

x₁ = 1/2 av²

for the distance it takes to stop, let's see what happens when we double the value v

x₂ = 1/2 A (2v)² = 4(1/2 * av²)

Now divide the formulae, getting

x₂ / x₁ = 

(4(1/2 * av²)) / (1/2 av²) = 4

Doing the division, everything cancels out except for the value 4. So if your speed doubles, the stopping distance quadruples, or becomes 4 times the original.

This is the answer: body 1 + engine 2

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