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Leno4ka [110]
3 years ago
10

Kelly met Jesse and Liza at the max for breakfast. The graph shows Kelly’s speed during her 12 minute drive

Mathematics
2 answers:
igomit [66]3 years ago
8 0

Answer:

Step-by-step explanation:

skelet666 [1.2K]3 years ago
3 0

Answer:

D. -360\,\frac{km}{h}

Step-by-step explanation:

Physically speaking, average acceleration (\bar a), measured in kilometers per square hour, is defined by following formula:

\bar a = \frac{v_{f}-v_{o}}{t_{f}-t_{o}} (1)

Where:

v_{o}, v_{f} - Initial and final speed, measured in kilometers per hour.

t_{o}, t_{f} - Initial and final instant, measured in hours.

The initial and final instants are, respectively:

t_{o} = 5\,min\times \frac{1\,h}{60\,min}

t_{o} = 0.083\,h

t_{f} = 9\,min \times \frac{1\,h}{60\,min}

t_{f} = 0.15\,h

If we know that t_{o} = 0.083\,h, v_{o} = 60\,\frac{km}{h}, t_{f} = 0.15\,h and v_{f} = 36\,\frac{km}{h}, then the average acceleration of Kelly is:

\bar a = \frac{36\,\frac{km}{h}-60\,\frac{km}{h}}{0.15\,h-0.083\,h}

\bar a = -358.209\,\frac{km}{h^{2}}

The correct answer is D.

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Generally, the equation for the problem is  mathematically given as

&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}

\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}

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