3.3 Equations of motion and free fall

Achievable AP Physics 1

3. Kinematics

Our AP Physics 1 course is currently in development and is a work-in-progress.

Equations of motion and free fall

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In this sub-chapter we discuss:

Three constant-acceleration equations
Graphical derivations
Algebraic derivations
Connection to kinematic graphs
Free fall scenarios

First equation of motion

When acceleration is constant, the average acceleration equals the change in velocity divided by the elapsed time:

$a = \frac{v - v _{0}}{t},$

where

$v_{0}$ is the initial velocity,
$v$ is the final velocity, and
$t$ is the elapsed time.

Solve this equation for $v$ by multiplying both sides by $t$ and then adding $v_{0}$ :

$v = v_{0} + a t .$

Below is a velocity-time graph for constant acceleration. The straight line shows that velocity changes linearly with time, and the slope of the line equals the acceleration.

Velocity-time graph for const. acceleration

Second equation of motion

For constant acceleration, the average velocity over a time interval is the mean of the initial and final velocities:

$v_{avg} = \frac{v _{0} + v}{2}$

Displacement over time $t$ is average velocity times time:

$Δ s = v_{avg} \cdot t$

Now substitute $v = v_{0} + a t$ into the average-velocity formula:

$v_{avg} = \frac{v _{0} + ( v _{0} + a t )}{2} = v_{0} + \frac{1}{2} a t .$

Substitute this into the displacement equation:

$Δ s = (v_{0} + \frac{1}{2} a t) t = v_{0} t + \frac{1}{2} a t^{2} .$

Since $Δ s = s - s_{0}$ , the position at time $t$ is:

$s = s_{0} + v_{0} t + \frac{1}{2} a t^{2} .$

Below is a position-time graph for constant acceleration. The curve is parabolic because of the quadratic term $\frac{1}{2} a t^{2}$ . The vertical difference between any two points on the curve gives the displacement between those times.

Position-time graph for const. acceleration

Third equation of motion

Sometimes you want a relationship that doesn’t include time. To eliminate $t$ , start with:

$v = v_{0} + a t .$

and the displacement equation:

$Δ s = v_{0} t + \frac{1}{2} a t^{2} .$

Solve the velocity equation for $t$ :

$t = \frac{v - v _{0}}{a} .$

Substitute this into the displacement equation:

$Δ s = v_{0} (\frac{v - v _{0}}{a}) + \frac{1}{2} a (\frac{v - v _{0}}{a})^{2} .$

Simplifying gives:

$v^{2} = v_{0}^{2} + 2 a Δ s,$

or equivalently,

$v^{2} = v_{0}^{2} + 2 a (s - s_{0}) .$

Even though the derivation above is algebraic, a velocity-time graph helps with interpretation: displacement is the area under the velocity-time curve, and this equation links that displacement to the change from $v_{0}$ to $v$ .

Example problem 1

A car starts from rest and accelerates uniformly at $2 m/s^{2}$ for a time $t_{1}$ . Then it decelerates uniformly at $3 m/s^{2}$ until coming to a stop. If the total time of the trip is $10 s$ , determine:

(a) The time $t_{1}$ spent accelerating.
(b) The maximum speed reached by the car.
(c) The total displacement during the trip.

Solution:

(spoiler)

(a) Let $t_{2} = 10 - t_{1}$ be the deceleration time.

During acceleration, the final speed is $v_{max} = 2 t_{1}$ .

During deceleration, the car slows from $v_{max}$ to 0 with deceleration $3 m/s^{2}$ , so:

$v_{max} = 3 t_{2} = 3 (10 - t_{1}) .$

Set the two expressions for $v_{max}$ equal:

$2 t_{1} = 3 (10 - t_{1}) \Rightarrow 2 t_{1} = 30 - 3 t_{1} \Rightarrow 5 t_{1} = 30 \Rightarrow t_{1} = 6 s .$

Thus, $t_{2} = 4 s$ .

(b) Maximum speed is:

$v_{max} = 2 t_{1} = 2 (6) = 12 m/s .$

(c) Displacement during acceleration:

$s_{1} = \frac{1}{2} (2) t_{1}^{2} = t_{1}^{2} = 36 m .$

Displacement during deceleration (using $v_{max} = 3 t_{2}$ ):

$s_{2} = v_{max} t_{2} - \frac{1}{2} (3) t_{2}^{2} = 12 \cdot 4 - \frac{1}{2} \cdot 3 \cdot 16 = 48 - 24 = 24 m .$

Total displacement:

$s = s_{1} + s_{2} = 36 + 24 = 60 m .$

The area under the graph will also give a total displacement of 60 m.

Example problem 2

An object’s position is described by:

$s (t) = 5 + 8 t - 2 t^{2}, t \in [0, 4] .$

Determine:

(a) The velocity at $t = 3 s$ .
(b) The displacement between $t = 0$ s and $t = 4$ s.
(c) The average velocity over this interval.

Solution:

(spoiler)

Write the position function in the standard form:

$s (t) = s_{0} + v_{0} t + \frac{1}{2} a t^{2} .$

Comparing, we have $s_{0} = 5$ , $v_{0} = 8$ , and $\frac{1}{2} a = - 2$ so that $a = - 4 m/s^{2}$ .

Velocity at $t = 3$ : Using $v = v_{0} + a t$ :

$v (3) = 8 + (- 4) (3) = 8 - 12 = - 4 m/s .$

The negative sign indicates that the object is moving in the opposite direction to the defined positive direction.
Displacement:

$Δ s = s (4) - s (0) .$

Calculate:

$s (0) = 5, s (4) = 5 + 8 \cdot 4 - 2 \cdot 16 = 5 + 32 - 32 = 5 m .$

So, $Δ s = 5 - 5 = 0 m$ ; the object returns to its starting position.
Average velocity: Since the net displacement is 0, the average velocity is:

$v_{avg} = \frac{0}{4} = 0 m/s .$

Position-time graph for the given scenario

Example problem 3

A sprinter starts from rest and accelerates uniformly during the first 3 seconds of a race. If the sprinter covers 12 m during this time, determine his final speed at 3 s. (Assume constant acceleration.)

Solution:

(spoiler)

Since the sprinter starts from rest, $s_{0} = 0$ and $v_{0} = 0$ .
The displacement for constant acceleration is given by:

$s = \frac{1}{2} a t^{2} .$

Given $s = 12$ m and $t = 3$ s:

$12 = \frac{1}{2} a (3^{2}) \Rightarrow 12 = \frac{9}{2} a \Rightarrow a = \frac{24}{9} \approx 2.67 m/s^{2} .$

The final speed is then:

$v = v_{0} + a t = 0 + 2.67 \times 3 \approx 8.0 m/s .$

Free fall

Free fall is a special case of constant acceleration where the only force acting is gravity. If we take upward as positive, then the acceleration is negative:

$a = - g (with g \approx 9.8 m/s^{2}) .$

Substitute $a = - g$ into the kinematic equations:

$v s v^{2} = v_{0} - g t, = s_{0} + v_{0} t - \frac{1}{2} g t^{2}, = v_{0}^{2} - 2 g (s - s_{0}) .$

These equations let you analyze free-fall motion using the same constant-acceleration tools, with the sign coming from your choice of positive direction.

Position-time graph for an object thrown upwards

Example problem 4 A ball is dropped from the top of a 45 m high building. Assuming $g = 9.8 m/s^{2}$ (take $g \approx 10 m/s^{2}$ for simplicity if desired), determine:

(a) The time taken for the ball to hit the ground.
(b) The speed of the ball just before impact.

Solution:

(spoiler)

(a) In free fall from rest, the displacement is given by:

$s = s_{0} - \frac{1}{2} g t^{2} .$

Assuming $s_{0} = 45$ m and $s = 0$ (ground level),

$0 = 45 - \frac{1}{2} (9.8) t^{2} \Rightarrow \frac{1}{2} (9.8) t^{2} = 45.$

Solving for $t$ :

$t^{2} = \frac{45 \times 2}{9.8} \approx \frac{90}{9.8} \approx 9.18 \Rightarrow t \approx 3.03 s .$

(b) The final speed is given by:

$v = g t \approx 9.8 \times 3.03 \approx 29.7 m/s .$

Thus, the ball takes approximately 3.03 s to fall and hits the ground at about 29.7 m/s.

Position-time graph for the given free fall scenario

Example problem 5 A ball is thrown upward with an initial speed of $20 m/s$ from the top of a $30 m$ building. Determine:

(a) The maximum height reached above the building.
(b) The total time of flight until the ball hits the ground.

Solution:

(spoiler)

(a) For upward motion under gravity, use the equation:

$v = v_{0} - g t .$

At the maximum height, $v = 0$ , so:

$0 = 20 - 9.8 t_{up} \Rightarrow t_{up} \approx \frac{20}{9.8} \approx 2.04 s .$

The additional height gained above the building is given by:

$Δ s = v_{0} t_{up} - \frac{1}{2} g t_{up}^{2} \approx 20 (2.04) - 0.5 (9.8) (2.04)^{2} .$

Calculating approximately:

$Δ s \approx 40.8 - 20.4 \approx 20.4 m .$

Thus, the maximum height above the ground is $30 + 20.4 \approx 50.4 m$ .

(b) The total time of flight can be found by considering the entire motion from the top of the building to the ground. Use the equation:

$0 = 30 + 20 t - \frac{1}{2} 9.8 t^{2} .$

Rewriting:

$4.9 t^{2} - 20 t - 30 = 0.$

Solve this quadratic using the quadratic formula:

$t = \frac{20 \pm 2 0 ^{2} + 4 \cdot 4.9 \cdot 30}{2 \cdot 4.9} .$

Calculate the discriminant:

$Δ = 400 + 588 = 988,$

so,

$t = \frac{20 + 988}{9.8} \approx \frac{20 + 31.4}{9.8} \approx \frac{51.4}{9.8} \approx 5.25 s .$

Thus, the total time of flight is approximately 5.25 s.

Three constant-acceleration equations

First: $v = v_{0} + a t$
Second: $s = s_{0} + v_{0} t + \frac{1}{2} a t^{2}$
Third: $v^{2} = v_{0}^{2} + 2 a (s - s_{0})$

Graphical and algebraic derivations

Velocity-time graph: slope = acceleration, area under curve = displacement
Position-time graph: parabolic curve for constant $a$ (due to $t^{2}$ term)
Algebraic manipulation connects equations and eliminates variables as needed

Connection to kinematic graphs

Slope of $v$ - $t$ graph = acceleration
Area under $v$ - $t$ graph = displacement
$s$ - $t$ graph shape: parabola for constant acceleration

Free fall scenarios

Special case: $a = - g$ (downward), $g \approx 9.8 m/s^{2}$
Kinematic equations with $a = - g$ :
- $v = v_{0} - g t$
- $s = s_{0} + v_{0} t - \frac{1}{2} g t^{2}$
- $v^{2} = v_{0}^{2} - 2 g (s - s_{0})$
Sign convention: upward positive, downward negative

Example problem 1 (car accelerates and decelerates)

Time accelerating: $t_{1} = 6 s$ , decelerating: $t_{2} = 4 s$
Maximum speed: $12 m/s$
Total displacement: $60 m$

Example problem 2 (object with quadratic position function)

Velocity at $t = 3 s$ : $- 4 m/s$ (moving opposite to positive direction)
Displacement $t = 0$ to $t = 4 s$ : $0 m$ (returns to start)
Average velocity: $0 m/s$

Example problem 3 (sprinter from rest)

Acceleration: $a \approx 2.67 m/s^{2}$
Final speed at $t = 3 s$ : $8.0 m/s$

Example problem 4 (ball dropped from building)

Time to hit ground: $\approx 3.03 s$
Speed just before impact: $\approx 29.7 m/s$

Example problem 5 (ball thrown upward from building)

Maximum height above ground: $\approx 50.4 m$
Total time of flight: $\approx 5.25 s$

Problem-solving strategy

Choose the kinematic equation that matches knowns and unknowns
For free fall, keep sign convention consistent (up positive, $a = - g$ )

Next | 3.4 Motion in two dimensions

Equations of motion and free fall

In this sub-chapter we discuss:

Three constant-acceleration equations
Graphical derivations
Algebraic derivations
Connection to kinematic graphs
Free fall scenarios

First equation of motion

When acceleration is constant, the average acceleration equals the change in velocity divided by the elapsed time:

$a = \frac{v - v _{0}}{t},$

where

$v_{0}$ is the initial velocity,
$v$ is the final velocity, and
$t$ is the elapsed time.

Solve this equation for $v$ by multiplying both sides by $t$ and then adding $v_{0}$ :

$v = v_{0} + a t .$

Below is a velocity-time graph for constant acceleration. The straight line shows that velocity changes linearly with time, and the slope of the line equals the acceleration.

Second equation of motion

For constant acceleration, the average velocity over a time interval is the mean of the initial and final velocities:

$v_{avg} = \frac{v _{0} + v}{2}$

Displacement over time $t$ is average velocity times time:

$Δ s = v_{avg} \cdot t$

Now substitute $v = v_{0} + a t$ into the average-velocity formula:

$v_{avg} = \frac{v _{0} + ( v _{0} + a t )}{2} = v_{0} + \frac{1}{2} a t .$

Substitute this into the displacement equation:

$Δ s = (v_{0} + \frac{1}{2} a t) t = v_{0} t + \frac{1}{2} a t^{2} .$

Since $Δ s = s - s_{0}$ , the position at time $t$ is:

$s = s_{0} + v_{0} t + \frac{1}{2} a t^{2} .$

Third equation of motion

Sometimes you want a relationship that doesn’t include time. To eliminate $t$ , start with:

$v = v_{0} + a t .$

and the displacement equation:

$Δ s = v_{0} t + \frac{1}{2} a t^{2} .$

Solve the velocity equation for $t$ :

$t = \frac{v - v _{0}}{a} .$

Substitute this into the displacement equation:

$Δ s = v_{0} (\frac{v - v _{0}}{a}) + \frac{1}{2} a (\frac{v - v _{0}}{a})^{2} .$

Simplifying gives:

$v^{2} = v_{0}^{2} + 2 a Δ s,$

or equivalently,

$v^{2} = v_{0}^{2} + 2 a (s - s_{0}) .$

Example problem 1

(a) The time $t_{1}$ spent accelerating.
(b) The maximum speed reached by the car.
(c) The total displacement during the trip.

Solution:

(spoiler)

(a) Let $t_{2} = 10 - t_{1}$ be the deceleration time.

During acceleration, the final speed is $v_{max} = 2 t_{1}$ .

During deceleration, the car slows from $v_{max}$ to 0 with deceleration $3 m/s^{2}$ , so:

$v_{max} = 3 t_{2} = 3 (10 - t_{1}) .$

Set the two expressions for $v_{max}$ equal:

$2 t_{1} = 3 (10 - t_{1}) \Rightarrow 2 t_{1} = 30 - 3 t_{1} \Rightarrow 5 t_{1} = 30 \Rightarrow t_{1} = 6 s .$

Thus, $t_{2} = 4 s$ .

(b) Maximum speed is:

$v_{max} = 2 t_{1} = 2 (6) = 12 m/s .$

(c) Displacement during acceleration:

$s_{1} = \frac{1}{2} (2) t_{1}^{2} = t_{1}^{2} = 36 m .$

Displacement during deceleration (using $v_{max} = 3 t_{2}$ ):

$s_{2} = v_{max} t_{2} - \frac{1}{2} (3) t_{2}^{2} = 12 \cdot 4 - \frac{1}{2} \cdot 3 \cdot 16 = 48 - 24 = 24 m .$

Total displacement:

$s = s_{1} + s_{2} = 36 + 24 = 60 m .$

The area under the graph will also give a total displacement of 60 m.

Example problem 2

An object’s position is described by:

$s (t) = 5 + 8 t - 2 t^{2}, t \in [0, 4] .$

Determine:

(a) The velocity at $t = 3 s$ .
(b) The displacement between $t = 0$ s and $t = 4$ s.
(c) The average velocity over this interval.

Solution:

(spoiler)

Write the position function in the standard form:

$s (t) = s_{0} + v_{0} t + \frac{1}{2} a t^{2} .$

Comparing, we have $s_{0} = 5$ , $v_{0} = 8$ , and $\frac{1}{2} a = - 2$ so that $a = - 4 m/s^{2}$ .

Velocity at $t = 3$ : Using $v = v_{0} + a t$ :

$v (3) = 8 + (- 4) (3) = 8 - 12 = - 4 m/s .$

The negative sign indicates that the object is moving in the opposite direction to the defined positive direction.
Displacement:

$Δ s = s (4) - s (0) .$

Calculate:

$s (0) = 5, s (4) = 5 + 8 \cdot 4 - 2 \cdot 16 = 5 + 32 - 32 = 5 m .$

So, $Δ s = 5 - 5 = 0 m$ ; the object returns to its starting position.
Average velocity: Since the net displacement is 0, the average velocity is:

$v_{avg} = \frac{0}{4} = 0 m/s .$

Example problem 3

Solution:

(spoiler)

Since the sprinter starts from rest, $s_{0} = 0$ and $v_{0} = 0$ .
The displacement for constant acceleration is given by:

$s = \frac{1}{2} a t^{2} .$

Given $s = 12$ m and $t = 3$ s:

$12 = \frac{1}{2} a (3^{2}) \Rightarrow 12 = \frac{9}{2} a \Rightarrow a = \frac{24}{9} \approx 2.67 m/s^{2} .$

The final speed is then:

$v = v_{0} + a t = 0 + 2.67 \times 3 \approx 8.0 m/s .$

Free fall

Free fall is a special case of constant acceleration where the only force acting is gravity. If we take upward as positive, then the acceleration is negative:

$a = - g (with g \approx 9.8 m/s^{2}) .$

Substitute $a = - g$ into the kinematic equations:

$v s v^{2} = v_{0} - g t, = s_{0} + v_{0} t - \frac{1}{2} g t^{2}, = v_{0}^{2} - 2 g (s - s_{0}) .$

These equations let you analyze free-fall motion using the same constant-acceleration tools, with the sign coming from your choice of positive direction.

Example problem 4 A ball is dropped from the top of a 45 m high building. Assuming $g = 9.8 m/s^{2}$ (take $g \approx 10 m/s^{2}$ for simplicity if desired), determine:

(a) The time taken for the ball to hit the ground.
(b) The speed of the ball just before impact.

Solution:

(spoiler)

(a) In free fall from rest, the displacement is given by:

$s = s_{0} - \frac{1}{2} g t^{2} .$

Assuming $s_{0} = 45$ m and $s = 0$ (ground level),

$0 = 45 - \frac{1}{2} (9.8) t^{2} \Rightarrow \frac{1}{2} (9.8) t^{2} = 45.$

Solving for $t$ :

$t^{2} = \frac{45 \times 2}{9.8} \approx \frac{90}{9.8} \approx 9.18 \Rightarrow t \approx 3.03 s .$

(b) The final speed is given by:

$v = g t \approx 9.8 \times 3.03 \approx 29.7 m/s .$

Thus, the ball takes approximately 3.03 s to fall and hits the ground at about 29.7 m/s.

Example problem 5 A ball is thrown upward with an initial speed of $20 m/s$ from the top of a $30 m$ building. Determine:

(a) The maximum height reached above the building.
(b) The total time of flight until the ball hits the ground.

Solution:

(spoiler)

(a) For upward motion under gravity, use the equation:

$v = v_{0} - g t .$

At the maximum height, $v = 0$ , so:

$0 = 20 - 9.8 t_{up} \Rightarrow t_{up} \approx \frac{20}{9.8} \approx 2.04 s .$

The additional height gained above the building is given by:

$Δ s = v_{0} t_{up} - \frac{1}{2} g t_{up}^{2} \approx 20 (2.04) - 0.5 (9.8) (2.04)^{2} .$

Calculating approximately:

$Δ s \approx 40.8 - 20.4 \approx 20.4 m .$

Thus, the maximum height above the ground is $30 + 20.4 \approx 50.4 m$ .

(b) The total time of flight can be found by considering the entire motion from the top of the building to the ground. Use the equation:

$0 = 30 + 20 t - \frac{1}{2} 9.8 t^{2} .$

Rewriting:

$4.9 t^{2} - 20 t - 30 = 0.$

Solve this quadratic using the quadratic formula:

$t = \frac{20 \pm 2 0 ^{2} + 4 \cdot 4.9 \cdot 30}{2 \cdot 4.9} .$

Calculate the discriminant:

$Δ = 400 + 588 = 988,$

so,

$t = \frac{20 + 988}{9.8} \approx \frac{20 + 31.4}{9.8} \approx \frac{51.4}{9.8} \approx 5.25 s .$

Thus, the total time of flight is approximately 5.25 s.

Achievable AP Physics 1

3. Kinematics

Our AP Physics 1 course is currently in development and is a work-in-progress.