Ask Question
11 November, 08:17

Juan starts walking at 4 miles per hour. An hour and a half later, Cathy starts jogging along the same route at 7 miles per hour. How long will it take Cathy to catch up with Juan?

+1
Answers (2)
  1. 11 November, 09:37
    0
    The answer to your question is 2 hours

    Step-by-step explanation:

    Data

    Juan speed = 4 mi/h

    Cathy speed = 7 mi/h

    t = x

    Formula

    speed = distance/time

    Solve for distance

    distance = speed x time

    d = s x t

    Distance travelled by Juan

    d = 4t

    Distance travelled by Juan in 1.5 h

    d = 4 (1.5)

    d = 6 mi

    total distance d = 6 + 4t

    Distance travelled by Cathy

    d = vt

    d = 7t

    Equal both equations

    6 + 4t = 7t

    Solve for t

    6 = 7t - 4t

    6 = 3t

    6/3 = t

    2 = t

    Cathy wll catch up with Juan in 2 hours
  2. 11 November, 11:59
    0
    Answer: it take Cathy 3.75 to catch up with Juan

    Step-by-step explanation:

    Let t represent the time it will take for Cathy to catch up with Juan. At time t hours when Cathy will catch up Juan, they would have covered the same distance.

    Juan starts walking at 4 miles per hour.

    Distance = speed *, time

    This means that the distance that Juan covered in to hours would be

    4 * t = 4t

    An hour and a half later, Cathy starts jogging along the same route at 7 miles per hour. Total time spent by Cathy would be t - 1.5

    Distance covered by Cathy would be

    7 (t - 1.5) = 7t - 10.5

    Since the distance covered is the sane, the.

    4t = 7t - 10.5

    7t - 4t = 10.5

    3t = 10.5

    t = 10.5/3 = 3.5 hours
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Juan starts walking at 4 miles per hour. An hour and a half later, Cathy starts jogging along the same route at 7 miles per hour. How long ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers