4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

• Position in straight-line motion is denoted by s(t)
• Velocity is the rate of change of position with respect to time.
• Velocity is calculated as the first derivative of the position function.
• Acceleration is calculated as the first derivative of the velocity function.
• Acceleration can also be calculated as the second derivative of the position function.
• Match the concept with its definition and mathematical representation:

  Position ↔️ Location at time $t$ ||| $s(t)$

  Velocity ↔️ Rate of change of position ||| $v(t) =$$s'(t)$

  Acceleration ↔️ Rate of change of velocity ||| $a(t) =$$v'(t) =$$s''(t)$
• Order the steps for solving problems involving straight-line motion concepts.

  1️⃣ Define position, velocity, and acceleration

  2️⃣ Understand relationships through derivatives

  3️⃣ Apply calculus to find unknowns

  4️⃣ Interpret results
• Motion graphs can be used to analyze the relationships between position, velocity, and acceleration.
• What does velocity measure in straight-line motion?

  Rate of change of position
• In straight-line motion, the location of an object along a line at time $t$ is called its position
• Acceleration in straight-line motion is the rate of change of velocity with respect to time.
• What is the mathematical representation of velocity in straight-line motion?

  $v(t) =$$s'(t)$
• The location of an object along a line at time $t$ is referred to as its position
• The acceleration in straight-line motion is given by $a(t) =$$v'(t) =$$s''(t)$.
• What does the position function $s(t)$ represent in straight-line motion?


  Object's location
• What is the mathematical representation of acceleration in straight-line motion?

  $a(t) =$$v'(t) =$$s''(t)$
• In straight-line motion, velocity is the rate of change of position
• The acceleration in straight-line motion is equal to the second derivative of the position function.
• What is the definition of acceleration in straight-line motion?

  Rate of change of velocity
• In straight-line motion, the velocity $v(t)$ is the first derivative of the position function.
• What does the derivative $s'(t)$ represent in straight-line motion?


  Velocity
• The acceleration in straight-line motion is calculated as $a(t) =$$v'(t) =$$s''(t)$.
• In straight-line motion, the derivative $s'(t)$ represents the velocity
• What does the second derivative $s''(t)$ represent in straight-line motion?


  Acceleration
• The first derivative of velocity gives acceleration in straight-line motion.
• To find the velocity function $v(t)$ from a position function $s(t)$, you take the derivative
• If $s(t) =$$2t^{2} +$$5t - 3$, what is the velocity function $v(t)$?


  $4t +$$5$
• Positive velocity in straight-line motion indicates forward motion.
• How is the acceleration function a(t)</latex> calculated from the velocity function $v(t)$?


  $a(t) =$$v'(t)$
• If $v(t) =$$4t +$$5$, the acceleration function is 4
• Constant acceleration means the velocity changes at a constant rate.
• Steps to find velocity and acceleration from position in straight-line motion:

  1️⃣ Find the first derivative of position to get velocity

  2️⃣ Find the first derivative of velocity to get acceleration
• What is the mathematical relationship between velocity and position?

  $v(t) =$$s'(t)$
• The velocity $v(t)$ is the first derivative of the position
• The derivative of position with respect to time gives velocity.
• If $s(t) =$$3t^{2} +$$5t - 2$, what is the velocity functionv(t)</latex>?


  $6t +$$5$
• What is the second derivative of position with respect to time called?

  Acceleration
• If an object's position is given by $s(t) =$$3t^{2} +$$2t - 1$, its velocity is v(t) = 6t + 2</latex>, and its acceleration is $a(t) =$$6$. The object's acceleration is constant at 6
• In straight-line motion, velocity is the rate of change of an object's position with respect to time.
• What is the mathematical relationship between velocity and position in straight-line motion?

  $v(t) =$$s'(t)$