Athletic competition

In a 400 meter athletic competition, a participant covers the distance as given below.
find the average speed?

first 80 meters 10 m/s
next 240 meters 7.5 m/s
last 80 meters 10 m/s

Correct result:

v =  8.333 m/s

Solution:

s=vt s1=80 m s2=240 m s3=80 m  v1=10 m/s v2=7.5 m/s v3=10 m/s  t1=s1/v1=80/10=8 s t2=s2/v2=240/7.5=32 s t3=s3/v3=80/10=8 s  v=s1+s2+s3t1+t2+t3=80+240+808+32+8=253=8.333 m/ss=vt \ \\ s_{1}=80 \ \text{m} \ \\ s_{2}=240 \ \text{m} \ \\ s_{3}=80 \ \text{m} \ \\ \ \\ v_{1}=10 \ \text{m/s} \ \\ v_{2}=7.5 \ \text{m/s} \ \\ v_{3}=10 \ \text{m/s} \ \\ \ \\ t_{1}=s_{1}/v_{1}=80/10=8 \ \text{s} \ \\ t_{2}=s_{2}/v_{2}=240/7.5=32 \ \text{s} \ \\ t_{3}=s_{3}/v_{3}=80/10=8 \ \text{s} \ \\ \ \\ v=\dfrac{ s_{1}+s_{2}+s_{3} }{ t_{1}+t_{2}+t_{3} }=\dfrac{ 80+240+80 }{ 8+32+8 }=\dfrac{ 25 }{ 3 }=8.333 \ \text{m/s}



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