Motorist drive

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 stopped track 35 m.

Correct answer:

v = 16.7554 m/s

Step-by-step explanation:

t_{0} = 0.8 s s = 35 m a = 6.5 m/s^{2} a = v / t t = v / a s = v \cdot t_{0} + \frac{1}{2} \cdot a \cdot t_{1}^{2} s = v \cdot t_{0} + \frac{1}{2} \cdot a \cdot (v / a)^{2} s = v \cdot t_{0} + 0.5 \cdot a \cdot (v / a)^{2} 35 = v \cdot 0.8 + 0.5 \cdot 6.5 \cdot (v / 6.5)^{2} - 0.07692307692308 v^{2} - 0.8 v + 35 = 0 0.07692307692308 v^{2} + 0.8 v - 35 = 0 a = 0.076923; b = 0.8; c = - 35 D = b^{2} - 4 a c = 0 . 8^{2} - 4 \cdot 0.076923 \cdot (- 35) = 11.4092307692 D > 0 v_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 0 . 8 \pm 1 1 . 4 1}{0 . 1 5 3 8 4 6} v_{1, 2} = - 5.2 \pm 21.955409 v_{1} = 16.755409356 v_{2} = - 27.155409356 v = v_{1} = 16.7554 ≐ 16.7554 m/s V = v \to km/h = v \cdot 3.6 km/h = 16.7554 \cdot 3.6 km/h = 60.31947 km/h Verifying Solution: t_{1} = v / a = 16.7554 / 6.5 ≐ 2.5778 s S = v \cdot t_{0} + \frac{1}{2} \cdot a \cdot t_{1}^{2} = 16.7554 \cdot 0.8 + \frac{1}{2} \cdot 6.5 \cdot 2.577 8^{2} = 35 m

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