A jeepney from rest
accelerates uniformly over a time of 3.25 seconds and covers a distance of 15
meters. What is the given initial velocity?
To find the initial velocity of the jeepney, we can use the following kinematic equation:
\[v = u + at\]
where:
- \(v\) = final velocity (assuming the jeepney starts from rest, so final velocity is the velocity after 3.25 seconds)
- \(u\) = initial velocity
- \(a\) = acceleration
- \(t\) = time
Given that the distance covered is 15 meters, we can also use the equation:
\[s = ut + \frac{1}{2}at^2\]
Rearranging the above equation to solve for initial velocity:
\[u = \frac{s - \frac{1}{2}at^2}{t}\]
Given:
- \(s\) = 15 meters
- \(a\) = acceleration is the rate of change of velocity = change in velocity/time taken = final velocity - initial velocity / time taken
- \(t\) = 3.25 seconds
First, let's find the acceleration using the first kinematic equation:
\[a = \frac{v - u}{t}\]
Since the Jeepney accelerates uniformly from rest, the final velocity will be equal to the distance covered divided by time:
\[v = \frac{s}{t}\]
So,
\[a = \frac{\frac{s}{t} - u}{t}\]
Given \(s = 15\) meters and \(t = 3.25\) seconds:
\[a = \frac{\frac{15}{3.25} - u}{3.25}\]
\[a = \frac{4.615 - u}{3.25}\]
Now, we can substitute the value of acceleration into the equation to find the initial velocity:
\[u = \frac{15 - \frac{1}{2}(\frac{4.615 - u}{3.25})(3.25)^2}{3.25}\]
\[u = \frac{15 - \frac{1}{2}(4.615 - u)}{3.25}\]
\[u = \frac{15 - 2.3075 + u}{3.25}\]
\[3.25u = 12.6925\]
\[u = 3.9 \, \text{m/s}\]
Therefore, the given initial velocity of the Jeepney is \(3.9 \, \text{m/s}\).