Answer to Question #23367 in Mechanics | Relativity for jasmyne

Question

John goes for a run. From his house, he jogs north for exactly 5.0 min at an average speed of 8.0 km/h. He continues north at a speed of 12.0 km/h for the next 30.0 min. He then turns around and jogs south at a speed of 15.0 km/h for 15.0 min. Then he jogs south for another 20.0 min at 8.0 km/h. He walks the rest of the way home.

How many kilometers does John jog in total?

Answer 1

1. As John jogs north for exactly 5.0 min at an& average speed of 8.0 km/h, he jogged

D1 = 5.0[m]·8.0[km/h] = 1/12[h]·8.0[km/h] = 2/3[km]

north.

2. As John continued north at a speed of 12.0 km/h for the next 30.0 min, he moved

D2 = 30.0[m]·12.0[km/h] = 1/2[h]·12.0[km/h] = 6[km]

north.

3. As John jogs south at a speed of 15.0 km/h for 15.0 min, he moved

D3 = 15.0[m]·15.0[km/h] = 1/4[h]·15.0[km/h] = 3.75[km]

south.

4. As John jogs south for another 20.0 min at 8.0 km/h, he moved

D4 = 20.0[m]·8.0[km/h] = 1/3[h]·8.0[km/h] = 8/3[km]

south.

Therefore, John moved

DN = D1 + D2 = 2/3[km] + 6[km] = 20/3[km]

north and

DS = D3 + D4 = 3.75[km] + 8/3[km] = 77/12[km]

south. So, he needs to walk

D = |20/3[km] -& 77/12[km]| = 3/12[km]

to return home. But the question is about how many kilometers does John jog in total, so the answer is

D = DS + DN = 20/3[km] + 77/12[km] = 157/12[km].