A ball is rolled down a hill with an initial velocity of 4.2m/s. How did it roll in 11.6s if it accelerates at 2.3m/s2
To determine how far the ball rolled down the hill, we can use the equation of motion:
\(d = v_0t + \frac{1}{2}at^2\)
Where:
- \(d\) is the distance traveled
- \(v_0\) is the initial velocity (4.2 m/s)
- \(t\) is the time taken (11.6 s)
- \(a\) is the acceleration (2.3 m/s²)
Plugging in the given values:
\(d = (4.2 \, \text{m/s})(11.6 \, \text{s}) + \frac{1}{2}(2.3 \, \text{m/s²})(11.6 \, \text{s})^2\)
\(d = 48.72 \, \text{m} + 0.5 \times (2.3 \, \text{m/s²}) \times (11.6 \, \text{s})^2\)
\(d = 48.72 \, \text{m} + 0.5 \times 2.3 \, \text{m/s²} \times 134.56 \, \text{s}^2\)
\(d = 48.72 \, \text{m} + 0.5 \times 308.888 \, \text{m}\)
\(d = 48.72 \, \text{m} + 154.444 \, \text{m}\)
\(d = 203.164 \, \text{m}\)
Therefore, the ball rolled a distance of 203.164 meters down the hill in 11.6 seconds.