Sprinter

Sprinter runs the relay 4 x 400 m to the handover at a speed of 42 km/h. A second runner is at the start of the handover area 20 m long and runs when it is the first sprinter at a distance of 10 m. Calculate the speed at which the second runner must run to the handover that occurred at the end of the handover area. Consider that the speed of both runners is constant.

Correct answer:

v =  28 km/h

Step-by-step explanation:

s_{1} = 20 \ \text{m} \rightarrow \ \text{km} = 20 : 1000 \ \ \text{km} = 0.02 \ \text{km} \ \\ s_{2} = 10 \ \text{m} \rightarrow \ \text{km} = 10 : 1000 \ \ \text{km} = 0.01 \ \text{km} \ \\ v_{2} = 42 \ \text{km/h} \ \\ \ \\ v = s/t \ \\ t_{1} = t_{2} \ \\ \ \\ v/s_{1} = v_{2}/(s_{1}+s_{2}) \ \\ v/0.02 = 42/(0.02+0.01) \ \\ \ \\ 50v = 1400 \ \\ \ \\ v = \dfrac{ 1400 }{ 50 } = 28 \ \\ \ \\ v = #km/h#s1 = m2km(20) \ \\ s2 = m2km(10) \ \\ v2 = 42 !km/h \ \\ \ \\ @v = s/t \ \\ @t1 = t2 \ \\ \ \\ l: v/s1 = v2/(s1+s2)v = 28 = 28 \ \text{km/h}



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