In an historical movie, two knights on horseback start from rest 94.3 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.273 m/s2, while Sir Alfred's has a magnitude of 0.280 m/s2. Relative to Sir George's starting point, where do the knights collide?
To find the collision point between the two knights, we need to first determine the time it takes for them to collide. We can use the kinematic equation:
\[d = ut + \frac{1}{2}at^2\]
where:
- \(d\) is the distance covered
- \(u\) is the initial velocity
- \(t\) is the time taken
- \(a\) is the acceleration
Let's start with Sir George. Since he starts from rest, his initial velocity is 0. Sir George's distance from the collision point is 94.3 m. We can plug these values into the equation:
\[94.3 = 0 \cdot t + \frac{1}{2} \cdot 0.273 \cdot t^2 \]
Simplifying the equation gives us:
\[0.1365t^2 = 94.3\]
Now, we can solve for \(t\):
\[t^2 = \frac{94.3}{0.1365}\]
\[t^2 \approx 691.11\]
\[t \approx \sqrt{691.11}\]
\[t \approx 26.30\]
So, it takes Sir George approximately 26.30 seconds to reach the collision point.
Next, let's calculate the distance Sir Alfred covers during this time. We can use the equation:
\[d = ut + \frac{1}{2}at^2\]
For Sir Alfred, his initial velocity is also 0, and he covers a distance of \(d = 94.3\) m, but his acceleration is 0.280 m/s². Plugging in these values:
\[94.3 = 0 \cdot 26.30 + \frac{1}{2} \cdot 0.280 \cdot 26.30^2\]
Simplifying the equation gives us:
\[0.0736t^2 = 94.3\]
Solving for \(t\) again:
\[t^2 = \frac{94.3}{0.0736}\]
\[t^2 \approx 1280.71\]
\[t \approx \sqrt{1280.71}\]
\[t \approx 35.80\]
So, it takes Sir Alfred approximately 35.80 seconds to reach the collision point.
Now that we have the time it takes for each knight to reach the collision point, we can calculate the distance each knight covers. We can use the equation:
\[d = ut + \frac{1}{2}at^2\]
Let's start with Sir George:
\[d_{George} = 0 \cdot 26.30 + \frac{1}{2} \cdot 0.273 \cdot (26.30)^2\]
\[d_{George} \approx 94.3\]
So, relative to Sir George's starting point, the collision point is approximately 94.3 meters away.
Note: Since Sir George and Sir Alfred have the same starting point, the collision point will be the same relative to both knights' starting positions.