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

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+82538.33338.333 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 } \doteq \dfrac{ 25 }{ 3 } \doteq 8.3333 \doteq 8.333 \ \text{m/s}



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