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3 years ago

Mrs. Johnson put 22,250 miles

Mathematics

1 answer:

Viefleur [7K]3 years ago

7 0

The change in quantities can be represented as percentage

<em>The percentage decrease in miles driven is 34.83%</em>

The given parameters are:

$\mathbf{Initial = 22250}$

$\mathbf{Final = 14500}$

The percentage decrease is calculated as:

$\mathbf{\%Decrease = \frac{Final - Initial}{Initial} \times 100\%}$

Substitute known values

$\mathbf{\%Decrease = \frac{14500- 22250}{22250} \times 100\%}$

$\mathbf{\%Decrease = \frac{-7750}{22250} \times 100\%}$

$\mathbf{\%Decrease = \frac{-775000}{22250} \%}$

$\mathbf{\%Decrease = -34.83 \%}$

The negative sign implies that, the change represents a decrement.

Hence, the percentage decrease in miles driven is 34.83%

Read more about percentage change at:

brainly.com/question/14801224

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

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The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

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To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

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$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

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$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

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Answer:

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