Two cars, Car A and Car B, travel along a straight highway. At time $t = 0\text{ h}$, Car A starts from position $s_A(0) = 15\text{ km}$ with position function $s_A(t) = 15 + 4(12t + 5)\text{ km}$, where $t$ is measured in hours. Car B starts from position $s_B(0) = 5\text{ km}$ and moves according to $s_B(t) = 5 + 7(8t - 3) + 2\int_0^t (3\tau + 10)\,d\tau\text{ km}$ at low speeds, which simplifies over constant cruise velocity to $s_B(t) = 5 + 7(8t - 3) + 26t\text{ km}$. Determine the exact time $t$ in hours when Car A and Car B meet at the same coordinate position.

Two cars, Car A and Car B, travel along a straight highway. At time $t = 0\text{ h}$ , Car A starts from position $s_A(0) = 15\text{ km}$ with position function $s_A(t) = 15 + 4(12t + 5)\text{ km}$ , where $t$ is measured in hours. Car B starts from position $s_B(0) = 5\text{ km}$ and moves according to $s_B(t) = 5 + 7(8t - 3) + 2\int_0^t (3\tau + 10)\,d\tau\text{ km}$ at low speeds, which simplifies over constant cruise velocity to $s_B(t) = 5 + 7(8t - 3) + 26t\text{ km}$ . Determine the exact time $t$ in hours when Car A and Car B meet at the same coordinate position.

Final answer

$t = 1.5\text{ h}$

Steps

Simplify the position function for Car A

To find the time $t$ when Car A and Car B meet, set their position functions equal to each other: $s_A(t) = s_B(t)$

Use the distributive property to expand the expression for $s_A(t)$ :

$s_A(t) = 15 + 4(12t + 5)$ $s_A(t) = 15 + 48t + 20$

Combine the constant terms:

$s_A(t) = 48t + 35$

Simplify the position function for Car B

Expand the brackets and combine like terms in $s_B(t)$ :

$s_B(t) = 5 + 7(8t - 3) + 26t$ $s_B(t) = 5 + 56t - 21 + 26t$

Combine the variable terms and constant terms:

$s_B(t) = (56t + 26t) + (5 - 21)$ $s_B(t) = 82t - 16$

Set the two equations equal and solve for $t$

Equate the simplified expressions:

$48t + 35 = 82t - 16$

Collect the variable terms on one side and the constant terms on the other side by subtracting $48t$ from both sides:

$35 = 82t - 48t - 16$ $35 = 34t - 16$

Add $16$ to both sides:

$35 + 16 = 34t$ $51 = 34t$

Divide both sides by $34$ :

$t = \frac{51}{34}$

Divide the numerator and denominator by their greatest common factor, $17$ :

$t = \frac{3}{2} = 1.5\text{ h}$

Pitfall to avoid: When distributing numbers across parentheses with subtraction, such as $7(8t - 3)$ , ensure the negative sign is applied correctly to get $-21$ , not $+21$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the multiplication to the constant inside parentheses when expanding 4(12t + 5), yielding 48t + 5 instead of 48t + 20.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
S_A(t) = S_B(t)
Step 2
15 + 48t + 5 = 5 + 56t - 21 + 26t
Correct
15 + 48t + 20 = 5 + 56t - 21 + 26t
When expanding $4(12t + 5)$ , you multiplied $4 \times 12t = 48t$ , but forgot to multiply the $4$ by the second term $5$ , leaving $+5$ instead of $+20$ .
Step 3
48t + 20 = 82t - 16
Correct
48t + 35 = 82t - 16
Your combining of like terms on both sides was algebraically consistent with your previous line, but it inherited the earlier expansion error.
Step 4
36 = 34t
Correct
51 = 34t
You correctly added $16$ and subtracted $48t$ across the equals sign based on your previous equation.
Step 5
t = $\frac{36}{34}$ = $\frac{18}{17} $\text{ h}$
Correct
t = $\frac{51}{34}$ = 1.5 $\text{ h}$
You accurately simplified the fraction $\frac{36}{34}$ , but the numerical value was affected by the initial distribution error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly established the equation $s_A(t) = s_B(t)$ to represent the two cars meeting.

You expanded the brackets on the right-hand side $7(8t - 3)$ accurately as $56t - 21$ .

You combined like terms and isolated the variable terms using correct equation-solving techniques.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete Distributive Property

You distributed the multiplier $4$ to the variable term $12t$ but left the constant $+5$ unmultiplied inside the parentheses.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Partial distribution across parentheses occurs when attention focuses on the variable term, causing the outer factor to be dropped before multiplying the constant term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Double-check every term in parentheses: For any expression $a(b + c)$ , the factor $a$ must multiply every term inside: $a \cdot b + a \cdot c$ .
Draw distribution arcs from the outer number to each individual term inside the parentheses before writing down the expanded expression.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the distributive property fully: $15 + 4(12t) + 4(5) = 15 + 48t + 20$ .
Combine constant terms on the left: $48t + 35 = 82t - 16$ .
Collect variable terms on one side and constants on the other: $35 + 16 = 82t - 48t \implies 51 = 34t$ .
Solve for $t$ : $t = \frac{51}{34} = \frac{3}{2} = 1.5\text{ h}$ .

Final answer

t = 1.5 $\text{ h}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice expanding linear binomials by explicitly writing out both multiplication steps.
Perform a mental check on the units and constant terms after expanding parentheses to confirm the multiplier reached all terms.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when expanding 7(8t - 3) by writing + 21 instead of - 21, resulting in an incorrect constant term for Car B.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
s_A(t) = s_B(t) $\implies$ 15 + 48t + 20 = 5 + 56t + 21 + 26t
Correct
s_A(t) = s_B(t) $\implies$ 15 + 48t + 20 = 5 + 56t - 21 + 26t
When distributing $7$ across the expression $(8t - 3)$ , multiplying $7$ by $-3$ yields $-21$ , not $+21$ .
Step 2
48t + 35 = 82t + 26
Correct
48t + 35 = 82t - 16
You correctly combined like terms based on your previous line ( $5 + 21 = 26$ ), but because of the initial sign error, the constant term on the right became $26$ instead of $5 - 21 = -16$ .
Step 3
34t = 9
Correct
34t = 51
You correctly subtracted $48t$ and $26$ across the equals sign ( $82t - 48t = 34t$ and $35 - 26 = 9$ ), but the constant on the left should be $35 - (-16) = 51$ .
Step 4
t = $\frac{9}{34} $\text{ h}$
Correct
t = $\frac{51}{34}$ = 1.5 $\text{ h}$
You divided both sides correctly by $34$ , but the resulting value reflects the earlier sign mistake.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly equated the two position functions $s_A(t) = s_B(t)$ to model when the two cars meet.

You successfully expanded the left side $4(12t + 5)$ into $48t + 20$ .

You maintained consistent, algebraically sound steps when combining like terms and isolating the variable.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Property Sign Error

You wrote $+21$ instead of $-21$ when expanding $7(8t - 3)$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the subtraction sign inside parentheses when distributing a factor, treating $a(b - c)$ as $ab + ac$ instead of $ab - ac$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Preserve Negative Signs During Distribution: $a(b - c) = a \cdot b - a \cdot c = ab - ac$
Always treat subtraction inside parentheses as addition of a negative number: $7(8t + (-3)) = 7(8t) + 7(-3) = 56t - 21$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the equation: $15 + 48t + 20 = 5 + 56t - 21 + 26t$
Combine constant and variable terms on each side: $48t + 35 = 82t - 16$
Collect terms with $t$ on one side and constants on the other: $35 + 16 = 82t - 48t \implies 51 = 34t$
Solve for $t$ : $t = \frac{51}{34} = \frac{3}{2} = 1.5$

Final answer

t = 1.5 $\text{ h}$

NEXT STEPS

Final Takeaways from Your Teacher

Circle or highlight negative signs inside brackets before multiplying through.
Verify your distributed terms by factoring them back out to check signs.
Complete the targeted flashcards and practice drills below to reinforce sign accuracy during distribution.

OverviewACEpath Teacher's Diagnosis

03 The student confuses the positions with velocities and equates the rates of change without accounting for the initial positions and constant offsets.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
v_A = 48 $\implies$ v_B = 56 + 26 = 82
Step 2
48t = 82t
Correct
48t + 35 = 82t - 16
You set the positions equal as $48t = 82t$ , which assumes both cars start at position $0$ . The full position expressions include constant terms: $s_A(t) = 15 + 4(12t + 5) = 48t + 35$ and $s_B(t) = 5 + 7(8t - 3) + 26t = 82t - 16$ .
Step 3
t = 0 $\text{ h}$
Correct
34t = 51 $\implies$ t = 1.5 $\text{ h}$
This result follows directly from the missing constant terms in your equation.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly calculated the velocity of Car A from the coefficient of $t$ as $48\text{ km/h}$ .

You correctly combined the rate terms $56 + 26 = 82\text{ km/h}$ for Car B.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Initial Value/Constant Offset

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated meeting at the same coordinate position as having traveled equal distances from the origin ( $v_A t = v_B t$ ), omitting the initial positions and constant terms that appear when fully expanding $s_A(t)$ and $s_B(t)$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Full Position Function: Objects meet when their coordinate positions are equal: $s_A(t) = s_B(t)$ , not merely their displacements from origin without offsets.
Always expand and collect both the variable terms and constant terms before setting functions equal.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand and simplify $s_A(t)$ : $s_A(t) = 15 + 48t + 20 = 48t + 35$ .
Expand and simplify $s_B(t)$ : $s_B(t) = 5 + 56t - 21 + 26t = 82t - 16$ .
Equate the two simplified positions: $48t + 35 = 82t - 16$ .
Rearrange to solve for $t$ : $35 + 16 = 82t - 48t \implies 51 = 34t \implies t = \frac{51}{34} = 1.5\text{ h}$ .

Final answer

t = 1.5 $\text{ h}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice identifying the constant term $s(0)$ or offset whenever a position function is given in non-expanded form.
Solve linear equations with constants and variables on both sides by isolating like terms on opposite sides.

Lock it in

Question

What does the distributive property state for an expression of the form $a(b + c)$ ?

Answer

$a(b + c) = ab + ac$ . The multiplier outside must be multiplied by every term inside the parentheses.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to distribute the multiplication to the constant inside parentheses when expanding 4(12t + 5), yielding 48t + 5 instead of 48t + 20.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student makes a sign error when expanding 7(8t - 3) by writing + 21 instead of - 21, resulting in an incorrect constant term for Car B.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student confuses the positions with velocities and equates the rates of change without accounting for the initial positions and constant offsets.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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