A ball starts from rest and rolls down a hill with uniform acceleration, traveling 200m during the second 4.8s of its motion. How far did it travel the first 4.8 seconds
To find out how far the ball traveled in the first 4.8 seconds, we need to use the equation for the distance covered during constant acceleration:
\(d = ut + \frac{1}{2}at^2\),
where:
\(d\) is the distance traveled,
\(u\) is the initial velocity (which is zero since the ball starts from rest),
\(a\) is the acceleration, and
\(t\) is the time.
Given that the ball starts from rest, the initial velocity \(u\) is zero. We also know that the time \(t\) is 4.8 seconds. So, the equation becomes:
\(d = \frac{1}{2}at^2\).
To find \(d\), we first need to determine the acceleration \(a\).
We are told that the ball has a uniform acceleration. If we assume that the uniform acceleration is constant throughout the motion, we can use the equation for average acceleration:
\(a = \frac{v - u}{t}\),
where:
\(a\) is the average acceleration,
\(v\) is the final velocity, and
\(u\) is the initial velocity.
Since the ball starts from rest, the initial velocity \(u\) is zero. We are also given that the distance traveled in the second 4.8 seconds is 200m. Using this information, we can determine the final velocity \(v\) during this time interval.
The equation for distance covered during constant acceleration can also be written as:
\(d = \frac{1}{2}(v + u)t\).
Since the initial velocity \(u\) is zero, the equation simplifies to:
\(d = \frac{1}{2}vt\).
Plugging in the known values, we have:
\(200 = \frac{1}{2}vt\).
Now, we can solve this equation for \(v\):
\(v = \frac{2d}{t} = \frac{2 \cdot 200}{4.8} = 83.33 \, \text{m/s}\) (rounded to two decimal places).
Now that we have the final velocity, we can calculate the average acceleration \(a\) during the second 4.8 seconds:
\(a = \frac{v - u}{t} = \frac{83.33 - 0}{4.8} = 17.36 \, \text{m/s}^2\) (rounded to two decimal places).
Finally, we can use the previously mentioned equation \(d = \frac{1}{2}at^2\) to find the distance traveled in the first 4.8 seconds:
\(d = \frac{1}{2} \cdot 17.36 \cdot (4.8)^2 = 198.34 \, \text{m}\) (rounded to two decimal places).
Therefore, the ball traveled approximately 198.34 meters in the first 4.8 seconds.