Question

2. At 1 pm, you leave Ellensburg at a constant speed of 75 miles per hour. At 2:30 pm, your

Answer 1

Here, we are required to determine how many hours your friend will drive in order to catch you.

(a) Your friend will have to drive 7 and a half hours inorder to catch you.

(b) You both will be 675 miles away from Ellensburg at that time.

If you leave at 1 pm; At 2:30pm;

• That is; 1 and a half hours after leaving; you must have covered a distance, d = 75 × 1.5
• d = 112.5miles.

Therefore, your position; S after 2:30pm is given by;

S(a) = 75t + 112.5 miles from Ellensburg.

For your friend; travelling at 90miles/hr;

• His position is given as; S(b) = 90 × t

(a) For your friend to catch you, you both must be in the same position;

• i.e S(a) = S(b)

75t + 112.5 = 90t

90t -75t = 112.5

t = 112.5/15

t = 7.5hours

(b) To determine how far you both are from Ellensburg; we can either evaluate:

S(b) = 90t or S(a) = 75t + 112.5

Therefore, By evaluating S(b) = 90t.

S(b) = 90 × 7.5

S(b) = 675miles from Ellensburg.

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Answer 2

My friend will drive 7 1/2 hours to catch me, and we'll both be 675 miles from Ellensburg by then.

Since at 1 pm, you leave Ellensburg at a constant speed of 75 miles per hour, and at 2:30 pm, your friend discovers that you have left your iPhone and knows you cannot live without it, and your friend jumps in a car and leaves Ellensburg at a constant speed of 90 miles per hour, to determine how many hours your friend drive in order to catch you, and how far from Ellensburg will you both be at that time, the following calculations must be performed:

• 2:30 - 1:00 = 1:30 = 1.5
• 75 x 1.5 = 112.5
• 90 - 75 = 15
• 112.5 / 15 = 7.5
• 7.5 x 90 = 675

Therefore, my friend will drive 7 1/2 hours to catch me, and we'll both be 675 miles from Ellensburg by then.

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