SAT Nonlinear Equations & Systems Question 134 | Free SAT Prep

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Question 134·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The kinetic energy $K$ (in joules) of an object with mass $m$ (in kilograms) and speed $v$ (in meters per second) is given by

K = \frac{1}{2}mv^2.

Object A has mass $8$ kilograms and positive speed $v$ . Object B has mass $2$ kilograms and speed $v+3$ meters per second. The two objects have the same kinetic energy.

What is the value of $v$ ?

For an equal-value problem involving a nonlinear formula, write the formula once for each situation and set the resulting expressions equal. Simplify to a polynomial equation, solve it by factoring when possible, and then use any context restriction, such as a speed being positive, to eliminate invalid solutions.

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Write both energy expressions

Use $K=\frac{1}{2}mv^2$ once for Object A and once for Object B. Be sure to substitute each object's mass and speed.

Use the equal-energy statement

Since the kinetic energies are equal, set the two expressions you found equal to each other. This will produce an equation involving only $v$ .

Check the meaning of speed

After solving the quadratic equation, use the fact that the problem describes $v$ as a positive speed to determine which solution is valid.

Desmos Guide

Graph the two kinetic-energy expressions

In Desmos, enter y=4x^2 and y=(x+3)^2. Here, $x$ represents $v$ , and the graphs show the kinetic energies of Objects A and B.

Find the intersections

Select the intersection points of the two graphs. Their $x$ -coordinates are the values of $v$ for which the objects have equal kinetic energy.

Use the positive-speed condition

One intersection has a negative $x$ -coordinate and one has a positive $x$ -coordinate. Because $v$ is stated to be positive, choose the positive intersection's $x$ -coordinate.

Step-by-step Explanation

Write an expression for each kinetic energy

For Object A,

K_A=\frac{1}{2}(8)v^2=4v^2.

For Object B,

K_B=\frac{1}{2}(2)(v+3)^2=(v+3)^2.

Use the equal-energy condition

The objects have the same kinetic energy, so set the expressions equal:

4v^2=(v+3)^2.

Expand and rearrange:

\begin{aligned} 4v^2&=v^2+6v+9\\ 3v^2-6v-9&=0. \end{aligned}

Factor and apply the speed condition

Factor the equation:

3(v^2-2v-3)=0

3(v-3)(v+1)=0.

Thus, $v=3$ or $v=-1$ . Since $v$ represents a positive speed, $v=-1$ is not valid. Therefore, $v=3$ , which is choice D.

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Desmos Guide

Step-by-step Explanation

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Step-by-step Explanation