Jogging versus walking

Question

Fred and Wilma like to go to Geometry Park. Fred likes to jog and Wilma lies to walk. At the park there is a circular track for jogging. There is a straight line walking path that passes through the center of the jogging circle track. If Fred and Wilma start at the same spot and Fred jogs once around the circle and Wilma walks along the straight walking path across the circle and back so she ends up at the same spot they started at, How much faster (in percent) must Fred jog around the circle than Wilma is walking so that they end up at the same spot they started at at the same time?

Answer 1

Fred walks along the circumference of the circular track. Wilma walks along the diameter 'd' of the track. So, first calculate the distance that each must walk.

D_W = 2*d

D_F = π*d (using equation for circumference)

You can also calculate D_W and D_F using the simple equation, distance D = speed V * time T

Then

Wilma D_W = 2*d = V_W * T_W ------ (1)

Fred D_F = π*d = V_F * T_F ------ (2)

Since Wilma and Fred cover the distance in the same time, therefore,

T_W = T_F

Substitute this into Eqns 1 and 2.

2*d = V_W * T ------ (3)

π*d = V_F * T ------ (4)

Divide eqn 4 by eqn 3 to eliminate the common factors, 'd' and π'T'.

π/2 = V_F/V_W

So Fred's Speed, V_F = 1.57 * V_W

This tells us that Fred must walk more than 1.5x faster than Wilma to make it back to the starting point at the same time as Wilma.

Percent excess speed relative to Wilma's speed

= 100* (V_F - V_W)/V_W

You can now calculate the answer. You should get 57%

Answer 2

Hello,

Here's how we can solve this. We start by defining some variables. d is the diameter of the circular track. D_F is the distance that Fred travels, and D_W is the distance that Wilma travels. r_F is Fred's speed, and r_W is Wilma's speed. t is the time that it takes them from start to end, and since they start together and end together, t is the same for both Fred and Wilma.

Fred travels once around the circular track, so his distance is:

D_F = π d

and his speed is:

r_F = D_F / t = π d / t

Wilma travels along the diameter to the other side then back, so her distance is:

D_W = 2d

and her speed is:

r_W = D_W / t = 2d / t

The percent change in their speeds is:

[ ( rF – rW ) / rW ] · 100

= [ ( π d / t – 2d / t ) / (2d / t) ] · 100

Multiplying top and bottom by t / d we get:

= [ ( π – 2 ) / 2 ] · 100

= ( π – 2 ) · 50

≈ 57.08%

Answer 3

The distance Fred will jog around the circle is 2∏r. The distance Wilma will walk across the circle and back is 2D or 4r, where D is the Diameter and r is the radius. Comparing the distances:

2∏r ∏

——— = ——— = 1.571

4r 2

The distance around the circle is 57.1 % longer than the distance across the circle and back. So, Fred must on average jog 57.1% faster than Wilma will walk to be able to meet her at the same spot at the same time they both started.