Asked by Josh
A 0.525 kg ball starts from rest and rolls down a hill with uniform acceleration, traveling 154 m during the second 10.0 s of its motion.
How far did it roll during the first 4.00 of motion?
I thought the answer was .2464 m, but it has been a couple of days, and I don't even remember how I got that anymore. :)
So s(t)=(1/2)at^2
Take s(20) - s(10)= 154
so (1/2)a(20)^2 - (1/2)a(10)^2 =154 or
a = (2*154)/(400-100)
Now you can find a. Put that in the equation and find s(4)
How far did it roll during the first 4.00 of motion?
I thought the answer was .2464 m, but it has been a couple of days, and I don't even remember how I got that anymore. :)
So s(t)=(1/2)at^2
Take s(20) - s(10)= 154
so (1/2)a(20)^2 - (1/2)a(10)^2 =154 or
a = (2*154)/(400-100)
Now you can find a. Put that in the equation and find s(4)
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