I still need help on my other question as well...
A runner hopes to complete the 10,000 m run in less than 30.0 min. AFter running at constant speed exactly 27.0 min there are still 1100 m to go. The runner must then accelerate at .20 s^-2 m for how many seconds in order to achieve the desired time?
ok well I foudn if you consider the first part of the run to have zero acceleration then you get 5.494 s^-1 m for the initial velocity for the second part of the race before the acceleration occurs...
so I'm trying to solve for t for the second part
reorginizing this equation
X = Xo + Vo t + 2^-1 a t^2
has more than one t in it and can't be taken out because there is a Vo and a acceleration
my only other choice is
a = t^-1 (V - Vo)
the final velocity can be assumed to be zero which it wouldn't be because most people would continue to run after the finish line but even if you did assume it to be zero then when you rearanged for time
t = a^-1 - Vo
I get a negative time...
Physics (Algebra Bit) - QUESTION, Thursday, August 20, 2009 at 8:44pm
thanks for help on other question =]
Physics (Algebra Bit) - QUESTION, Thursday, August 20, 2009 at 8:47pm
Is the answer to (c) correct by the way for the other problem I didn't know how fast it would be accelerating realtive to the larger block because the larger block would be accelerating more at 5.2 s^-1 m and the smaller block accelerates at 3.2 s^-1 m so I wasn't sure how much the smaller block is accelerating with reference to the larger block
Physics (Algebra Bit) - Anonymous, Sunday, October 14, 2012 at 7:47pm
An arrow is fired with a speed of 24.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1750 m/s2 and the block's acceleration has a magnitude of 450 m/s2