A is 100m away from headquarters. B is 8m away from headquarters. A runs twice as fast as B. How far away from headquarters will B be when A arrives at headquarters.
I drew a number line and did a jump of 16 for A and a jump of 8 for B. This brings A to 84m away and B is at HQ. Why B continues to run, I have no idea, but then I did a jump of 84m for A bringing A to HQ and then B moves 42m away from HQ. I've tried other ways and keep arriving at the same answer, but the correct answer is supposed to be 30...what am I doing wrong?
I bet B starts at 80, not 8 and runs at speed v
A runs at speed 2v
t = 100/2v
Bdistance run = v t = v(100/2v) = 50
so
B is still 80 - 50 = 30 meters away
I agree...thanks, Damon. I'm calling a typo on this question and not going to worry about it anymore.
To solve this problem, you need to consider the relative speeds and distances traveled by A and B. One way to approach it is by using proportions.
Let's say the time it takes for A to reach the headquarters is represented by 't' units of time. Since A runs twice as fast as B, we can say that B will take 2t units of time to reach the headquarters.
Now, let's consider the distances traveled. A is 100m away from the headquarters and runs twice as fast as B. So, when A reaches the headquarters, it will have traveled a distance of 2 times the distance covered by B.
Using the equation d = vt (where d is the distance, v is the velocity, and t is the time), we can set up two equations:
For A: 100 = v_A * t
For B: d_B = v_B * (2t)
Given that B is 8m away from the headquarters, we have: 8 = v_B * (2t)
Now, let's find the value of v_A. Since A runs twice as fast as B, we can write v_A = 2 * v_B.
Substituting this back into the equation for A, we have 100 = 2 * v_B * t.
Now, let's solve these equations simultaneously:
8 = v_B * (2t) ---> (equation 1)
100 = 2 * v_B * t ---> (equation 2)
Dividing equation 2 by 2, we have 50 = v_B * t.
Substituting the value of v_B * t from equation 1 into equation 2, we have:
50 = 2 * (8 / 2)
Solving the equation, we get the value of t as 2.
So, it takes B 2 units of time to reach the headquarters, and since it is traveling at a speed of v_B = 8m/2t = 8m/4 = 2m per unit of time, we can calculate the distance it covers during that time: d_B = v_B * (2t) = 2m/unit of time * (2 units of time) = 4m
Therefore, when A arrives at the headquarters, B will be 4m away from the headquarters, not at the headquarters as you mentioned in your explanation.
To obtain the correct answer of 30m, please double-check the given information or provide more specific details about the problem.