Is the nearest walk to Brownian motion approximately uniform?

Uniform Motion Problem. (explain please I am lost!)?

  • Answer:

    He ran 5 mph faster than he walked. So if his walking speed is v, his running speed is v + 5. He ran 8 miles at speed v + 5. So using d = v*t, the time to run the 8 miles was 8/(v+5) hours. He walked 6 miles at speed v. So the time to walk the 6 miles was 6/v hours. The total of these times was 2 hours: 8/(v+5) + 6/v = 2. That's the equation you have to solve. You can solve it by multiplying both sides by v(v+5) which will clear the fractions and leave you with a quadratic equation.

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He ran 5 mph faster than he walked. So if his walking speed is v, his running speed is v + 5. He ran 8 miles at speed v + 5. So using d = v*t, the time to run the 8 miles was 8/(v+5) hours. He walked 6 miles at speed v. So the time to walk the 6 miles was 6/v hours. The total of these times was 2 hours: 8/(v+5) + 6/v = 2. That's the equation you have to solve. You can solve it by multiplying both sides by v(v+5) which will clear the fractions and leave you with a quadratic equation.

Randy P

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