In a historical movie, two knights on horseback start from rest 62 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.17 m/s2, while Sir Alfred's has a magnitude of 0.30 m/s2. Relative to Sir George's starting point, where do the knights collide?
d=1/2 at^2
use that for both. YOu know the time is the same, so use a proportion
d1/d2=a1/a2
knowing d2+d2=62, you can solve it.
To determine where the knights collide, we can use the concepts of motion and the equations of kinematics. Let's break down the problem step by step:
Step 1: Find the time it takes for the knights to collide.
To find the time, we can use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
where \( s \) is the distance traveled, \( u \) is the initial velocity, \( t \) is the time, and \( a \) is the acceleration.
For Sir George:
\( s = 62 \, \text{m} \)
\( u = 0 \, \text{m/s} \) (since he starts from rest)
\( a = 0.17 \, \text{m/s}^2 \)
For Sir Alfred:
\( s = 0 \, \text{m} \) (since Sir Alfred starts from Sir George's position)
\( u = 0 \, \text{m/s} \) (since Sir Alfred starts from rest)
\( a = -0.30 \, \text{m/s}^2 \) (since Sir Alfred's acceleration is in the opposite direction)
Plugging these values into the equation, we have:
\( 62 = 0 \cdot t + \frac{1}{2} \cdot 0.17 \cdot t^2 \) (for Sir George)
\( 0 = 0 \cdot t + \frac{1}{2} \cdot (-0.30) \cdot t^2 \) (for Sir Alfred)
Simplifying, we get:
\( 0.085t^2 = 62 \) (for Sir George)
\( -0.15t^2 = 0 \) (for Sir Alfred)
To solve for \( t \), we can divide both sides of the equation by the coefficient of \( t^2 \), yielding:
\( t^2 = \frac{62}{0.085} \) (for Sir George)
\( t^2 = 0 \) (for Sir Alfred)
Since \( t^2 = 0 \) is a trivial case, we can ignore it. Solving the other equation, we find:
\( t = \sqrt{\frac{62}{0.085}} \approx 9.48 \, \text{s} \)
Step 2: Determine the distance traveled by Sir George and Sir Alfred during this time.
We can use the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
For Sir George:
\( u = 0 \, \text{m/s} \)
\( a = 0.17 \, \text{m/s}^2 \)
\( t = 9.48 \, \text{s} \)
For Sir Alfred:
\( u = 0 \, \text{m/s} \)
\( a = -0.30 \, \text{m/s}^2 \)
\( t = 9.48 \, \text{s} \)
Plugging these values into the equation, we have:
\( s_{\text{George}} = 0 \cdot 9.48 + \frac{1}{2} \cdot 0.17 \cdot (9.48)^2 \)
\( s_{\text{Alfred}} = 0 \cdot 9.48 + \frac{1}{2} \cdot (-0.30) \cdot (9.48)^2 \)
Simplifying, we find:
\( s_{\text{George}} \approx 7.60 \, \text{m} \)
\( s_{\text{Alfred}} \approx -17.02 \, \text{m} \)
Step 3: Determine the position of the collision relative to Sir George's starting point.
Since Sir Alfred starts from Sir George's position, the position of the collision relative to Sir George's starting point is simply the distance traveled by Sir Alfred: \( s_{\text{Alfred}} \approx -17.02 \, \text{m} \).
Therefore, the knights collide approximately 17.02 meters from Sir George's starting point, on Sir Alfred's side.