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willworkforfood
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Is it true that if the rate of decceleration is two times slower than the rate of acceleration for the same object, that it will need two times the distance to come to a stop as it did to accelerate?
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Doc Al said:Figure it out for yourself! How does the distance required to accelerate from a speed of 0 to V (or decelerate from V to 0) depend on the acceleration?
That's a start: You have a relation connecting acceleration and velocity (and time). Now figure out how to get the distance. (Hint: Distance equals average speed X time.)willworkforfood said:I know that the equation for acceleration is final velocity - initial velocity divided by time
You don't need actual values. You are trying to find the relationship between acceleration and distance. So far you have:willworkforfood said:The problem doesn't give me average speed or time though.
No... you need average speed, which is v/2, not v.willworkforfood said:So equation 2 would be x = v*t
When you combine the equations, don't eliminate v; instead, eliminate t. You want to end up with an equation relating v, a, and x.and you substitute v = at and get
x = a*t^2
Deceleration, also known as negative acceleration, is the rate at which an object's velocity decreases over time.
Deceleration is directly related to stopping distance. The greater the deceleration, the shorter the stopping distance will be.
Several factors affect deceleration, including the mass of the object, the force applied to it, and the friction between the object and the surface it is moving on.
Deceleration can be calculated by dividing the change in velocity by the time taken for the change to occur. It can also be calculated by dividing the force applied to an object by its mass.
The relationship between deceleration and stopping distance can be demonstrated through experiments using objects of different masses and applying different forces to them. The resulting stopping distances can be measured and compared to show the direct correlation between deceleration and stopping distance.