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)
To find the distance the ball rolled during the first 4.00 s of motion, you can use the equation for displacement:
s(t) = (1/2)at^2
First, let's find the acceleration (a). We are given that the ball traveled 154 m during the second 10.0 s of motion. So, we can set up the equation:
s(20) - s(10) = 154
Plug in the values:
(1/2)a(20)^2 - (1/2)a(10)^2 = 154
Simplifying the equation:
a(20)^2 - a(10)^2 = 308
Now, we can solve for a:
a(400) - a(100) = 308
400a - 100a = 308
300a = 308
a = 308/300
a ≈ 1.027 m/s^2
Now that we know the acceleration, we can substitute it back into the equation s(t) = (1/2)at^2 to find the distance traveled in the first 4.00 s.
s(4) = (1/2)(1.027)(4)^2
s(4) ≈ 0.2464 m
So, the ball rolled approximately 0.2464 m during the first 4.00 s of motion.