In a historical movie, two knights on horseback start from rest 64 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.21 m/s2, while Sir Alfred's has a magnitude of 0.31 m/s2. Relative to Sir George's starting point, where do the knights collide?
To determine where the knights collide, we need to find the time it takes for them to meet. We can do this by using the kinematic equation:
\[ d = v_i t + \frac{1}{2} a t^2 \]
where:
- \( d \) is the initial distance between the knights (64 m),
- \( v_i \) is the initial velocity (which is 0 since they start from rest),
- \( a \) is the acceleration of each knight, and
- \( t \) is the time it takes for the knights to meet.
First, let's find the time it takes for Sir George to reach the collision point. We can use the equation:
\[ d = \frac{1}{2} a t^2 \]
Plugging in the known values:
\[ 64 = \frac{1}{2} (0.21) t^2 \]
Simplifying the equation:
\[ t^2 = \frac{64}{0.21} \]
\[ t^2 \approx 304.76 \]
Taking the square root of both sides:
\[ t \approx \sqrt{304.76} \]
\[ t \approx 17.47 \]
Therefore, Sir George takes approximately 17.47 seconds to reach the collision point.
Now, let's find the time it takes for Sir Alfred to reach the collision point. Using the same equation:
\[ d = \frac{1}{2} a t^2 \]
Plugging in the known values:
\[ 64 = \frac{1}{2} (0.31) t^2 \]
Simplifying the equation:
\[ t^2 = \frac{64}{0.31} \]
\[ t^2 \approx 206.45 \]
Taking the square root of both sides:
\[ t \approx \sqrt{206.45} \]
\[ t \approx 14.36 \]
Therefore, Sir Alfred takes approximately 14.36 seconds to reach the collision point.
Since Sir Alfred reaches the collision point earlier, we can calculate the distance Sir Alfred travels before the collision:
\[ d = v_i t + \frac{1}{2} a t^2 \]
Plugging in the values:
\[ d = 0 \cdot 14.36 + \frac{1}{2} (0.31) (14.36)^2 \]
Simplifying the equation:
\[ d \approx 31.15 \]
Therefore, relative to Sir George's starting point, the knights collide approximately 31.15 meters away.