A soccer ball rolling across a field with an acceleration of -0.5m/s/s rolls 12.5 m and then comes to a stop after 8 seconds. What was the ball's initial speed?
V = Vo + a*t = 0
Vo -0.5*8 = 0
Vo = 4 m/s.
Should the distance rolled be 16 m?
Well, isn't that a ballin' problem! Let's kick things off by using the good old equation of motion:
v = u + at
Where:
v = final velocity (which is 0 because it comes to a stop)
u = initial velocity (what we're trying to find out)
a = acceleration (-0.5 m/s/s)
t = time (8 seconds)
Now, we can plug in the values and solve for u:
0 = u + (-0.5)(8)
Solving for u, it would be:
u = 0.5(8)
So, the ball's initial speed was a breathtaking 4 m/s!
To find the initial speed of the ball, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (zero in this case, as the ball comes to a stop)
u = initial velocity (what we want to find)
a = acceleration (-0.5 m/s^2, as given)
s = distance (12.5 m, as given)
Rearranging the equation, we have:
u^2 = v^2 - 2as
Substituting the known values:
u^2 = 0^2 - 2 * (-0.5 m/s^2) * 12.5 m
u^2 = 0 - (-12.5 m^2/s^2)
u^2 = 12.5 m^2/s^2
Taking the square root of both sides:
u = √(12.5 m^2/s^2)
Calculating the square root:
u ≈ 3.54 m/s
Therefore, the ball's initial speed was approximately 3.54 m/s.
To find the ball's initial speed, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (0 m/s, as the ball comes to a stop)
u is the initial velocity (what we want to find)
a is the acceleration (-0.5 m/s^2)
s is the distance traveled (12.5 m)
Rearranging the equation, we have:
u^2 = v^2 - 2as
Substituting the given values:
u^2 = (0 m/s)^2 - 2 * (-0.5 m/s^2) * (12.5 m)
Simplifying:
u^2 = 0 - (-12.5 m^2/s^2)
u^2 = 12.5 m^2/s^2
Taking the square root of both sides:
u = √(12.5 m^2/s^2)
Calculating the square root:
u ≈ 3.54 m/s
Therefore, the initial speed of the ball was approximately 3.54 m/s.