Knights' Collision: Solving a 1D Kinematics Problem for Homework

In summary: So, you can solve for X1 and X2.In summary, to find the distance from Sir George's starting point where the knights collide, you need to use equations that give the position of each knight relative to Sir George's starting position at any given time. By setting these equations equal to each other and using the fact that the total distance traveled by both knights is 62m, you can solve for the distance where they collide. This involves dividing one equation by the other and using the known values for acceleration for each knight.
  • #1
Neil6790
20
0

Homework Statement


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 accleration has a magnitude of 0.21 m/s2, while Sir Alfred's has a magnitude of 0.26 m/s2. How far from Sir George's starting point do the knights collide?

Homework Equations





The Attempt at a Solution


62 - x = (1/2)*(0.21 m/s^2)t^2
62 - x = (1/2)*(0.26 m/s^2)t^2

I was trying to find any velocity or any time but i can't seem to conceive anything logic that would work. Any help would be greatly appreciated.
 
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  • #2
Try to write equations that give the position of each knight relative to Sir George's starting position at any given time, and then find when they are equal.
 
  • #3
Neil6790 said:

Homework Statement


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 accleration has a magnitude of 0.21 m/s2, while Sir Alfred's has a magnitude of 0.26 m/s2. How far from Sir George's starting point do the knights collide?

Homework Equations



The Attempt at a Solution


62 - x = (1/2)*(0.21 m/s^2)t^2
62 - x = (1/2)*(0.26 m/s^2)t^2

I was trying to find any velocity or any time but i can't seem to conceive anything logic that would work. Any help would be greatly appreciated.

Almost the right idea.
You need to use different X1 and x2

X1 = 1/2a1t2

X2 = 1/2a2t2

You know that they meet by definition when the time is equal and ...their total distance adds to 62.

Divide the first equation by the second.

You get X1/X2 = a1/a2

and you know X1 + X2 = 62.
 

Related to Knights' Collision: Solving a 1D Kinematics Problem for Homework

1. What is 1D kinematics?

1D kinematics is the study of motion in one dimension, typically along a straight line. It involves analyzing the position, velocity, and acceleration of an object in motion.

2. How do you solve a simple 1D kinematics problem?

To solve a simple 1D kinematics problem, you will need to identify the given information, such as initial and final positions, velocities, and time. Then, you can use the equations of motion, such as the equation for displacement (Δx = v0t + 1/2at^2) or velocity (v = v0 + at), to solve for the unknown variable.

3. What are the three types of 1D motion?

The three types of 1D motion are constant velocity, constant acceleration, and free fall. In constant velocity, the object moves at a constant speed in a straight line. In constant acceleration, the object's velocity changes at a constant rate. In free fall, the object is only under the influence of gravity and accelerates at a constant rate of 9.8 m/s^2.

4. How do you calculate acceleration in a 1D kinematics problem?

To calculate acceleration, you can use the equation a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval. Alternatively, you can also use the equation a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time.

5. What is the difference between distance and displacement in 1D kinematics?

Distance is a scalar quantity that refers to the total length of the path traveled by an object, while displacement is a vector quantity that refers to the straight-line distance between the initial and final positions of an object. Displacement takes into account both the distance and direction of an object's motion, while distance only considers the distance traveled regardless of direction.

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