I still need help with this please.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?

Question

I still need help with this please.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?

Answer 1

I will gladly critique your work.

Answer 2

i don't understand the d1/d2=a1/a2.

Answer 3

d1=distance Sr Georg goes

d2= distance Sr alfred goes

a1= sr getorge's acceleration

d1/d2=a1/a2
but d2=62-d1
so

d1*a2=(62-d1)a1
solve for d1, then solve for d2

Answer 4

so you set it up like the following?

d1*.30=(62-d1).17 It has been several years since I had algebra and it should be common sense but how do I get the d1 by itself?

Answer 5

You are going to have a very difficult time in Physics, the question you are asking is algebra I.

.30d1+.17d1=62*.17
d1=62*.17 *1/(.47)
= 21m

Answer 6

To determine the location where the knights collide relative to Sir George's starting point, we need to analyze the motion of both knights and calculate the time it takes for them to collide.

The first step is to find the time it takes for each knight to collide. We can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance
u = initial velocity (which is 0 since they start from rest)
a = acceleration
t = time

For Sir George:
s = 62 m (the distance between the knights)
a = 0.17 m/s^2

Substituting the values into the equation, we get:

62 = 0 * t + (1/2) * 0.17 * t^2

Rearranging the equation, we have:

0.17 * t^2 = 62

Next, we solve for t:

t^2 = 62 / 0.17

t^2 ≈ 364.71

Taking the square root of both sides, we find:

t ≈ √364.71

t ≈ 19.10 seconds (rounded to two decimal places)

Now let's find the distance Sir George travels during this time. We can use the following equation:

s = ut + (1/2)at^2

Where:
s = distance
u = initial velocity (which is 0 since he starts from rest)
a = acceleration (0.17 m/s^2 from earlier)
t = time (19.10 seconds)

Substituting the values, we have:

s = 0 * 19.10 + (1/2) * 0.17 * (19.10)^2

s ≈ 0 + 0.17 * 182.71

s ≈ 31.04 meters (rounded to two decimal places)

Therefore, Sir George collides with Sir Alfred approximately 31.04 meters away from his starting point.