Question 140·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The stopping distance, $d$ , in meters for a certain car traveling at $v$ kilometers per hour on dry pavement is modeled by

d = 0.039v^2 + 0.1v.

Which of the following correctly expresses $v$ in terms of $d$ ?

A quadratic model has a squared speed term, so solving for speed can produce two branches. For formula choices, use Desmos to make one valid speed-distance pair, then check which expression returns the original speed. Ruling out every other choice settles the test, though one matching pair alone isn't a general proof. Watch the signs under the square root and before the linear coefficient.

Hints

Hint 1
An inverse formula runs the model backward: a speed gives you a stopping distance, and the inverse should return that same speed. Pick a convenient positive speed, such as $10$ kilometers per hour. What distance does the model give?
Hint 2
The radicand is the quantity under a square root. If it's negative, the expression has no real value. Once you have a valid distance, check the whole radicand in each proposed formula.
Hint 3
Substitute your distance into each proposed formula. An expression that doesn't return your starting speed can't undo the model. A match at one distance doesn't prove a formula works everywhere, but you can eliminate choices that fail.

Step-by-step

Approach 1: Test a speed-distance pair in Desmos
Step 1
Make a pair from the model
Choose $v=10$ kilometers per hour, a convenient positive speed. Type $0.039(10)^2+0.1(10)$ into Desmos; it prints $4.9$ . So the proposed inverse formula, which runs the model backward, must turn $d=4.9$ meters back into $v=10$ .
Step 2
Check the first radicand
Add $d=4.9$ , then type $\frac{0.1+\sqrt{0.01-0.156d}}{0.078}$ . Desmos shows undefined: the radicand, the number under the square root, is negative. That expression can't return the speed for this valid distance.
Step 3
Test the next expression
Type $\frac{-0.1+\sqrt{0.156-0.01d}}{0.078}$ . Desmos prints about $2.912$ , not the starting speed of $10$ . So this expression doesn't undo the model for your test pair.
Step 4
Test the positive-numerator expression
Type $\frac{0.1+\sqrt{0.01+0.156d}}{0.078}$ . Desmos prints about $12.564$ , not $10$ . The positive $0.1$ before the square root makes this result too large.
Step 5
Find the expression that returns the speed
Finally, type $\frac{-0.1+\sqrt{0.01+0.156d}}{0.078}$ . Desmos prints $10$ , the speed you started with. It is the only expression left after the other three fail. A known input-output pair can rule out proposed inverse formulas that don't return the input. So $v=\frac{-0.1+\sqrt{0.01+0.156d}}{0.078}$ gives the car's speed in terms of its stopping distance. Choice D.

Approach 2: Derive the formula with the quadratic formula
Step 1
Make one side zero
To use the quadratic formula, subtract $d$ from both sides: $0=0.039v^2+0.1v-d$ . Keep the minus sign on $d$ ; it controls the sign under the square root.
Step 2
Substitute the coefficients
Treat $d$ as fixed while solving for $v$ . In $av^2+bv+c=0$ , the coefficients are $a=0.039$ , $b=0.1$ , and $c=-d$ . The quadratic formula is $v=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . Substitute, keeping $-d$ in parentheses: $v=\frac{-0.1\pm\sqrt{(0.1)^2-4(0.039)(-d)}}{2(0.039)}$ .
Step 3
Square the linear coefficient
Square $0.1$ under the radical: $v=\frac{-0.1\pm\sqrt{0.01-4(0.039)(-d)}}{2(0.039)}$ . The constant $0.01$ comes from $b^2$ , so it doesn't get multiplied by $d$ .
Step 4
Simplify the distance term
Multiply $-4(0.039)(-d)$ ; two negative signs make it positive: $v=\frac{-0.1\pm\sqrt{0.01+0.156d}}{2(0.039)}$ .
Step 5
Simplify the denominator
Double $0.039$ in the denominator: $v=\frac{-0.1\pm\sqrt{0.01+0.156d}}{0.078}$ . The denominator divides the whole numerator, not only the square root.
Step 6
Choose the physical speed
A stopping distance has $d\ge0$ , so $\sqrt{0.01+0.156d}\ge0.1$ . The plus branch gives a nonnegative speed; the minus branch gives a negative one. The car's speed is $v=\frac{-0.1+\sqrt{0.01+0.156d}}{0.078}$ . Choice D.

Question 140·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

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Step-by-step

Lessons that teach this

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Strategy

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Step-by-step

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