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:

$$\begin{gathered} t_0=0.8\text{s} \\ s=35\text{m} \\ a=6.5{\text{m/s}}^2 \\ a=v/t \\ t=v/a \\ s=v\cdot t_0+\frac{1}{2}\cdot a\cdot t_1^2 \\ s=v\cdot t_0+\frac{1}{2}\cdot a\cdot (v/a{)}^2 \\ s=v\ast t_0+0.5\ast a\ast (v/a{)}^2 \\ 35=v\cdot 0.8+0.5\cdot 6.5\cdot (v/6.5{)}^2 \\ -0.076923v^2-0.8v+35=0 \\ 0.076923v^2+0.8v-35=0 \\ a=0.076923;b=0.8;c=-35 \\ D=b^2-4ac=0.8^2-4\cdot 0.076923\cdot (-35)=11.40922 \\ D>0 \\ {v}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-0.8\pm \sqrt{11.41}}{0.153846} \\ {v}_{1,2}=-5.2000052\pm 21.955420949746 \\ {v}_1=16.755415749741 \\ {v}_2=-27.155426149751 \\ \text{ Factored form of the equation: } \\ 0.076923(v-16.755415749741)(v+27.155426149751)=0 \\ v=v_1=16.7554\doteq 16.7554\text{m/s} \\ V=v\to \text{km/h}=v\cdot 3.6\text{km/h}=16.7554\cdot 3.6\text{km/h}=60.31947\text{km/h} \\ \text{ Verifying Solution: } \\ {t}_1=v/a=16.7554/6.5\doteq 2.5778\text{s} \\ S=v\cdot t_0+\frac{1}{2}\cdot a\cdot t_1^2=16.7554\cdot 0.8+\frac{1}{2}\cdot 6.5\cdot 2.5778^2=35\text{m} \end{gathered}$$

Our quadratic equation calculator calculates it.

Try another example