At what

At what speed did the motorist drive when he reacted by noticing the obstacle by 0.8 s; the magnitude of the opposite acceleration during braking was 6.5 m/s2 and the car ran to a stop track 35 m?

Correct answer:

v =  16.7554 m/s

Step-by-step explanation:

t0=0.8 s s=35 m a=6.5 m/s2  a=v/t t=v/a s=v t0+21 a t12 s=v t0+21 a (v/a)2  s=vt0+0.5a(v/a)2  35=v 0.8+0.5 6.5 (v/6.5)2 0.076923v20.8v+35=0 0.076923v2+0.8v35=0  a=0.076923;b=0.8;c=35 D=b24ac=0.8240.076923(35)=11.40922 D>0  v1,2=2ab±D=0.1538460.8±11.41 v1,2=5.2000052±21.955420949746 v1=16.755415749741 v2=27.155426149751   Factored form of the equation:  0.076923(v16.755415749741)(v+27.155426149751)=0  v=v1=16.755416.7554 m/s V=v km/h=v 3.6  km/h=16.7554 3.6  km/h=60.31947 km/h   Verifying Solution:   t1=v/a=16.7554/6.52.5778 s S=v t0+21 a t12=16.7554 0.8+21 6.5 2.57782=35 m

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