Failure

Drivers of passenger cars have calculated that at a speed of 60 km/h arrives at destination within 40 minutes. After 20 km refrain five minutes due a technical failure. How fast must go the rest of the way to the finish came at the scheduled time?

Result

v2 =  80 km/h

Solution:

v1=60 t=40/60=23≐0.6667 h s=v1⋅ t=60⋅ 0.6667=40 km s1=20 km  t1=s1/v1=20/60≐13≐0.3333 h s2=s−s1=40−20=20 km t2=t−t1−5/60=0.6667−0.3333−5/60=14=0.25 h v2=s2/t2=20/0.25=80 km/hv_{1}=60 \ \\ t=40/60=\dfrac{ 2 }{ 3 } \doteq 0.6667 \ \text{h} \ \\ s=v_{1} \cdot \ t=60 \cdot \ 0.6667=40 \ \text{km} \ \\ s_{1}=20 \ \text{km} \ \\ \ \\ t_{1}=s_{1}/v_{1}=20/60 \doteq \dfrac{ 1 }{ 3 } \doteq 0.3333 \ \text{h} \ \\ s_{2}=s - s_{1}=40 - 20=20 \ \text{km} \ \\ t_{2}=t - t_{1} - 5/60=0.6667 - 0.3333 - 5/60=\dfrac{ 1 }{ 4 }=0.25 \ \text{h} \ \\ v_{2}=s_{2}/t_{2}=20/0.25=80 \ \text{km/h}

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