Average speed

The average speed of a pedestrian who walked 10 km was 5km/h, the average speed of a cyclist on the same track was 20km/h. In how many minutes did the route take more than a cyclist? Q

Result

x =  90 min

Solution:

s=10 km v1=5 km/h v2=20 km/h  t=t1−t2 t1=s/v1=10/5=2 h t2=s/v2=10/20=12=0.5 h  t=t1−t2=2−0.5=32=1.5 h  x=t→min=t⋅ 60 min=90 min=90 mins=10 \ \text{km} \ \\ v_{1}=5 \ \text{km/h} \ \\ v_{2}=20 \ \text{km/h} \ \\ \ \\ t=t_{1}-t_{2} \ \\ t_{1}=s/v_{1}=10/5=2 \ \text{h} \ \\ t_{2}=s/v_{2}=10/20=\dfrac{ 1 }{ 2 }=0.5 \ \text{h} \ \\ \ \\ t=t_{1}-t_{2}=2-0.5=\dfrac{ 3 }{ 2 }=1.5 \ \text{h} \ \\ \ \\ x=t \rightarrow min=t \cdot \ 60 \ min=90 \ min=90 \ \text{min}

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