Motion2

The cyclist started off town at 19 km/h. The car started after 0.7 hours behind him in the same direction. It caught up with him for 23 minutes. How fast and how long went the car from the city to catch cyclists?

Correct answer:

v₂ = 53.7 km/h
s = 20.6 km

Step-by-step explanation:

v_{2} = 19 / 23 \cdot 60 \cdot (0.7 + 23 / 60) = \frac{1 2 3 5}{2 3} km/h = 53.7 km/h

s = 19 t = 19 \cdot (0.7 + \frac{2 3}{6 0}) = 20.6 km v_{2} = \frac{s}{\frac{2 3}{6 0}} = \frac{6 0 s}{2 3} = 53.7 k m / h

s = 19 t
s = v · 23/60
t = 0.7+23/60

s = 19·t
s = v · 23/60
t = 0.7+23/60

s-19t = 0
60s-23v = 0
60t = 65

Pivot: Row 1 ↔ Row 2
60s-23v = 0
s-19t = 0
60t = 65

Row 2 - 1/60 · Row 1 → Row 2
60s-23v = 0
-19t+0.383v = 0
60t = 65

Pivot: Row 2 ↔ Row 3
60s-23v = 0
60t = 65
-19t+0.383v = 0

Row 3 - -19/60 · Row 2 → Row 3
60s-23v = 0
60t = 65
0.383v = 20.583

v = 20.58333333/0.38333333 = 53.69565217
t = 65/60 = 1.08333333
s = 0+23v/60 = 0+23 · 53.69565217/60 = 20.58333333

s = 247/12 ≈ 20.583333
t = 13/12 ≈ 1.083333
v = 1235/23 ≈ 53.695652

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