Solving for Initial Velocity and Distance with an Accelerated Motion Formula
The given formula is:
\[ s = ut + \frac{1}{2}at^2 \]
To solve part (a), we want to make \( u \) the subject of the formula. Here's how we do it:
First, let's rearrange the formula to solve for \( u \) by subtracting \( \frac{1}{2}at^2 \) from both sides of the equation:
\[ s - \frac{1}{2}at^2 = ut \]
Next, we divide both sides of the equation by \( t \) to solve for \( u \):
\[ \frac{s - \frac{1}{2}at^2}{t} = u \]
So the formula with \( u \) as the subject is:
\[ u = \frac{s - \frac{1}{2}at^2}{t} \]
For part (b), we need to calculate \( s \) when \( a = 100 \text{ m/s}^2 \), \( u = 2 \text{ m/s} \), and \( t = 5 \text{ s} \). We can plug the given values into the original formula:
\[ s = ut + \frac{1}{2}at^2 \]
\[ s = (2 \text{ m/s})(5 \text{ s}) + \frac{1}{2}(100 \text{ m/s}^2)(5 \text{ s})^2 \]
\[ s = 10 \text{ m} + \frac{1}{2}(100 \text{ m/s}^2)(25 \text{ s}^2) \]
\[ s = 10 \text{ m} + 50(25 \text{ m}) \]
\[ s = 10 \text{ m} + 1250 \text{ m} \]
\[ s = 1260 \text{ m} \]
Thus, the distance \( s \) is 1260 meters.
