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(II) If the speed of a car is increased by $50 \% ,$ by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time.
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Physics 101 Mechanics
Work and Energy
University of Washington
Simon Fraser University
University of Sheffield
In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.
In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.
(II) If the speed of a car…
If the speed of a car is i…
When a driver brings a car…
A car initially going $50 …
What constant acceleration…
A car has a maximum accele…
The minimum stopping dista…
So here the initial velocity of the car is increased by 50%. And we're trying to find how the what factor does the minimum braking distance change? So we can say that if if the a massive the car massive em, we can draw a free bar dog body diagram for the car. So going straight down is the force of gravity going straight up would be forced normal, and going straight to the left would, of course, be force of friction. Ah, we could say that the work needed in order to ah, the work needed or to stop the car would be its strange and kinetic energy. And then we can we also know that the force of friction is going to be equal to the coefficient of kinetic friction times the normal force. And then we can say that some of forces in the white direction equals zero, which means that the force normal minus the force of gravity, equals zero, giving us that the force normal equals the force of gravity. So at this point, we can say that the only force in the ex direction that would be affecting work so force would be the only force Ah, force. The friction would be the only force that is ah, parallel to the distance to the displacement of the car. So we can say that the work done by friction which would in this case, be negative force of friction times the braking distance d equals 1/2 um v f squared minus V initial squared. So at this point, we know that the brakes were trying to stop the car so we can set the final velocity is going to be zero. And we can say that Ah, the coefficient of kinetic friction times thie forced normal. We also know that the forced normal is equal to the force of gravity, is weak and simply say force normal times. M g d equals 1/2. I am the initial squared. So at this point, we can say that the distance the braking distance is going to be directly proportional to the initial velocity squared. At this point, we can set up a relationship and we know that the braking distance over the original braking distance would be equal to B squared over the original Ah, initial velocity. So here and we know that the initial velocity is being increased by 50%. So we can say that V equals three times being natural, divided by two. And at this point, we can substitute, and we can say that to D is going to equal the original braking distance. Times three. The initial squared divided by two rather three. The initial divided by two quantity squared my apologies, and then it would be one times one rather one divided by V initial squared. Those would cancel out, and we had simply have. This would be equal to nine d nine times the original braking distance divided by four. So we can say that the distance will increase by a factor of 2.25 So the braking distance increases by a factor of 2.25 Given that the initial velocity is increased by 50% that is the end of the solution. Thank you for watching
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