Motion2

Cyclist started out of town at 19 km/h. After 0.7 hours car started behind him in the same direction and caught up with him for 23 minutes. How fast and how long went car from the city to caught cyclist?

Result

v2 =  53.7 km/h
s =  20.6 km

Solution:

v2=19/23⋅ 60⋅ (0.7+23/60)=53.7  km/h v_2=19/23 \cdot \ 60 \cdot \ (0.7+23/60) = 53.7 \ \text{ km/h }
s=19t=19⋅(0.7+2360)=20.6  km  v2=s2360=60s23=53.7 km/hs = 19 t = 19 \cdot (0.7+ \dfrac{ 23}{60}) = 20.6 \ \text{ km } \ \\ v_2 = \dfrac{ s } { \dfrac{ 23}{60} } = \dfrac{ 60 s } { 23 } = 53.7 \ km/h

s = 19 t
s = v * 23/60
t = 0.7+23/60

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

s = 24712 ≈ 20.583333
t = 1312 ≈ 1.083333
v = 123523 ≈ 53.695652

Calculated by our linear equations calculator.

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