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?
I have no idea where to start or how the distance is related to the acceleration.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