Kinematic problem, are my variables right,

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MarkFL said:
I simply used ##t## for the elapsed time which is fairly common practice, whereas your professor is having you use ##t_2-t_1##. It means the same thing, but your professor's notation is not unusual either.
lets say i was trying to do t2-t1 for x2 and and x1. so 2.5-6=-3.5. how does that work
 
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rashad764 said:
lets say i was trying to do t2-t1 for x2 and and x1. so 2.5-6=-3.5. how does that work

I'm not sure what you're asking here, but ##t##'s represent time and ##x##'s represent distance or position, so you don't want to interchange them.
 
MarkFL said:
I'm not sure what you're asking here, but ##t##'s represent time and ##x##'s represent distance or position, so you don't want to interchange them.
in this problem, what would be the time at position 2 and at position 1
 
rashad764 said:
in this problem, what would be the time at position 2 and at position 1

For the first phase of the problem, we could let:

##t_1=0\text{ s},\,t_2=6\text{ s}\implies t=\Delta t=v_2-v_1=6\text{ s}##

##x_1=0\text{ m},\,x_2=77.4\text{ m}\implies x=\Delta x=x_2-x_1=77.4\text{ m}##
 
thanks for helping, really appreciate it!
 
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