A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is initially traveling at speed 2v? Assume that the acceleration due to the braking is the same in both cases.

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Cherry McCormick · Accepted Answer

Step 1    We can find the acceleration from this kinematic equation:    vf2=vi2+2a(xf−xi)    Step 2    So when the car was traveling with initial speed:  vi=v    it takes a distance d to stop    Then:  xf−xi=d and vf=0    0=v2+2ad    −v2=2ad    a=−v22d    (1)    Step 3    When the car traveling with initial speed:  vi=2v    ∴PSK0=(2v)2+2ad2    −4v2=2ad2    d2=−4v22a    (2)    Notice that acceleration is the same in two cases.    Step 4    Plugging a from (1) into (2):    d2=−4v22a=4×−¬v2×¬2d2×−¬v2=4d

nick1337 · Accepted Answer

The distance traveled by the car can be determined using the equation:d=12at2+v0twhere:d is the distance traveled,a is the acceleration,t is the time taken,and v0 is the initial velocity.In the given problem, the car is initially traveling at a speed of v. Let&#039;s calculate the time it takes for the car to stop in this case. We know that the final velocity (vf) is 0 when the car comes to a stop. The equation relating final velocity, initial velocity, acceleration, and time is:vf=v0+atSince vf=0 when the car comes to a stop, we have:0=v−atSolving this equation for t, we get:t=vaNow, substitute this value of t into the equation for distance d:d=12a(va)2+v0(va)Simplifying, we have:d=v22a+v2ad=3v22aSo, the stopping distance when the car is initially traveling at speed v is 3v22a.Now, let&#039;s find the stopping distance when the car is initially traveling at speed 2v. We&#039;ll use the same equations of motion, but with v0 replaced by 2v. The time taken to stop will be:t=2vaSubstituting this value of t into the equation for distance d:d=12a(2va)2+2v(2va)Simplifying, we have:d=2v22a+4v2ad=5v2aTherefore, the stopping distance when the car is initially traveling at speed 2v is 5v2a.

Don Sumner · Accepted Answer

The equation of motion for the car can be written as:v2=u2+2aswhere u is the initial velocity, a is the acceleration, and s is the distance traveled.In this case, the car is initially traveling at speed v. So, we have:v2=v2+2adSimplifying this equation, we get:0=2adDividing both sides of the equation by 2a, we find:d=0Therefore, the stopping distance when the car is initially traveling at speed 2v is 0.

Vasquez · Accepted Answer

Result:3v22aSolution:When the car is traveling at speed 2v, we can consider the following equation of motion for the car during braking:v02=v2+2a·dvwhere v0 is the initial velocity (speed 2v), v is the final velocity (zero in this case), a is the acceleration due to braking, and dv is the stopping distance.Since the acceleration due to braking is the same in both cases, we can assume that a remains constant. Thus, we can rewrite the equation as:(2v)2=v2+2a·dvSimplifying this equation, we get:4v2=v2+2a·dvSubtracting v2 from both sides, we have:3v2=2a·dvNow, we can solve for dv by isolating it on one side of the equation:dv=3v22aTherefore, the stopping distance when the car is initially traveling at speed 2v is 3v22a.

A car traveling at speed v takes distance d to stop after the brakes are applied

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