Kinematics in One Dimension Problem

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A basketball player accelerates from rest to a speed of 6.0 m/s in 1.5 seconds, and the problem involves calculating the distance run during this time. The correct formulas to use are Vf = Vo + at and Vf^2 = Vo^2 + 2a(Delta x). The confusion arises from misidentifying 6.0 m/s as acceleration instead of velocity, which is crucial for accurate calculations. Understanding the distinction between velocity and acceleration is essential for solving kinematics problems effectively. Emphasizing conceptual understanding before applying formulas is recommended to avoid errors.
pstfleur
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1. In getting ready to slamdunk the ball, a basketball player starts from rest and sprints to a speed of 6.0m/s in 1.5seconds. Assuming that the player accelerates uniformly, determine the distance he runs



2. Vf=Vo+at, VF^2=Vo^2+2a(Delta x)



3. Vf= 0m/s+(6.0)(1.5)= 9
Delta X= Vf^2-V0^2/2a= 81-0/12=6.75

Delta X should be the distance he's running. The book answer is giving me 4.5m, but I am coming up with 6.75..Am I plugging in wrong values?
 
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pstfleur said:
3. Vf= 0m/s+(6.0)(1.5)= 9



6.0 m/s is a velocity not an acceleration. In other words, you need to find the acceleration,a. Vf=6 not a.
 
rock.freak667 said:
6.0 m/s is a velocity not an acceleration. In other words, you need to find the acceleration,a. Vf=6 not a.


Thanks.. That was the key.. Is there an easier way to understand which is velocity and which is acceleration. I am always having a hard time figuring out which is which.
 
pstfleur said:
... a basketball player starts from rest and sprints to a speed of 6.0m/s ...

Both, speed and the units m/s are keywords for velocity.

Before you start plugging away on formulae and doing math, wrap your head around the concepts of physics first. If you can't do that, you're bound for disaster.
 
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