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2 years ago

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At the end of a snow storm, Amelia saw there was a lot of snow on her front lawn. The temperature increased and the snow began t

o melt at a steady rate. After the storm, the snow started melting at a rate of 1.5 inches per hour and it is known that 4 hours after the storm ended, the depth of snow was down to 12 inches. Write an equation for S, in terms of t, representing the depth of snow on Amelia's lawn, in inches, t hours after the snow stopped falling.

1 answer:

Andrei [34K]2 years ago

5 0

Answer:

The equation is;

S = 18 - 1.5t

Step-by-step explanation:

From the question, we are made to know that the depth of the snow is S and the time taken to melt in hours is t

After 4 hours of melting, the snow depth is 12 inches

Now, recall that the rate of melting is 1.5 inches/ hour

So in the 4 hours, the amount of snow melted will be 1.5 * 4 = 6 inches

So the total depth of the snow initially will be 6 + 12 = 18 inches

So let’s write the equation;

S = 18 - 1.5t

Where S is the depth of the snow after a particular number of hours t

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Answer

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Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

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Where:

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$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

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We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

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<u>Josh:</u>

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

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Similarly solving for Josh we obtain

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Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

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$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

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<em>Whose textbook reaches the ground first and by how many seconds?</em>

Josh's textbook reached the ground first
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