Acceleration 7147

The runner ran the track 100m in 10.2 seconds. He ran the first 20m with a uniformly accelerated movement. Then he ran evenly. What was its acceleration on the 20m section, and what was its maximum speed?

Correct answer:

a =  3.4602 m/s2
v =  11.7647 m/s

Step-by-step explanation:

s=100 m s2=20 m t=10.2 s  s2=21 a t12 v=a t1  ss2=v (tt1)  2 s2=a t12=v t1 ss2=v (tt1)  ss2=tvvt1=t v2 s2  v=ts+s2=10.2100+20=1720011.7647 m/s  t1=2 s2/v=2 s2/17200=2 20/11.7647=517=3.4 s  a=v/t1=17200/517=17200 175=17 17200 5=2891000=2891000 m/s2=3.4602 m/s2
v=1720011.7647 m/s v2=v km/h=v 3.6  km/h=11.7647 3.6  km/h=42.353 km/h



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