CLASSICAL-MECHANICS · Interactive Practice | Unit 3 · Video 4

From Graph Areas to Kinematic Equations

IKey Formulas

Formula	Name	Description
$v(t) = v_0 + at$	First kinematic equation	Velocity is linear in time; slope $a$ , intercept $v_0$
$x(t) = x_0 + v_0 t + \tfrac{1}{2}a t^2$	Second kinematic equation	Displacement = area under the $v$ - $t$ graph
$\text{Area under } a\text{-}t = \Delta v$	Acceleration-area rule	Rectangle of height $a$ , width $t$ gives $\Delta v = at$
$v_{\text{ave}} = \dfrac{v_0 + v(t)}{2}$	Midpoint average velocity	Holds only for constant acceleration

IIInteractive Visualizations

Visualization 1 — Slope Is the Acceleration

On a velocity–time graph the intercept is $v_0$ and the slope is the constant acceleration $a$ .

Initial velocity v₀ 4 Acceleration a 2

Visualization 2 — Area Is the Displacement

The area under the velocity line is a rectangle $v_0 t$ plus a triangle $\tfrac{1}{2}a t^2$ — together the displacement $\Delta x$ .

Initial velocity v₀ 3 Acceleration a 1 Elapsed time t 6

💡 This trapezoid area is exactly the second kinematic equation $x(t) - x_0 = v_0 t + \tfrac{1}{2}a t^2$ .

Visualization 3 — Line Meets Parabola

A steady object's straight line leads at first, but an accelerating object's parabola overtakes it at $t_a = 2v_2/a_1$ .

Bus speed v₂ 1.5 Car acceleration a₁ 1 Now t 0

IIIQuiz Questions

Question 1

A body has initial velocity $v_0 = 5$ m/s and constant acceleration $a = 3$ m/s². Using the first kinematic equation $v(t) = v_0 + at$ , what is the velocity at $t = 4$ s?

✅ Correct! v = 5 + (3)(4) = 17 m/s.

❌ Not quite. Add the initial velocity to the velocity gained: v = v₀ + at = 5 + (3)(4).

Show solution

Solution:

Apply $v(t) = v_0 + at$ directly with $v_0 = 5$ , $a = 3$ , $t = 4$ :

$v(4) = 5 + (3)(4) = 5 + 12 = 17 \text{ m/s}$

The initial velocity is added to the velocity gained, $at = 12$ m/s.

Question 2

For constant acceleration, the area under the acceleration-time graph from $0$ to $t$ equals the change in velocity $\Delta v$ , not the displacement.

True or False?

✅ Correct! Area under a-t = a·t = Δv. The v-t area is what gives displacement.

❌ Not quite. The rectangle area a·t equals at = v(t) − v₀ = Δv. Displacement comes from the v-t graph instead.

Show solution

Solution:

The acceleration-time graph for constant $a$ is a horizontal line at height $a$ . Its area from $0$ to $t$ is a rectangle:

$\text{Area} = a \cdot t$

From the first kinematic equation, $at = v(t) - v_0 = \Delta v$ . So the area under the $a$ - $t$ graph equals the velocity change. (It is the area under the $v$ - $t$ graph that gives the displacement $\Delta x$ .)

The statement is True.

Question 3

A car starts from rest ( $v_0 = 0$ ) and accelerates uniformly, covering $100$ m while reaching $20$ m/s. Eliminating the acceleration between $x = \tfrac{1}{2}at^2$ and $v = at$ gives $t = 2x/v$ . How long did the acceleration phase last?

✅ Correct! t = 2x/v = 2(100)/20 = 10 s.

❌ Not quite. Use t = 2x/v with x = 100 m and v = 20 m/s, not t = x/v or t = v/a.

Show solution

Solution:

With $v_0 = 0$ , the equations simplify to $x = \tfrac{1}{2}at^2$ and $v = at$ . Substitute $a = v/t$ into the position equation:

$x = \tfrac{1}{2}\left(\tfrac{v}{t}\right)t^2 = \tfrac{1}{2}vt \;\;\Rightarrow\;\; t = \frac{2x}{v}$

Plug in $x = 100$ m and $v = 20$ m/s:

$t = \frac{2(100)}{20} = \frac{200}{20} = 10 \text{ s}$

(As a check, the acceleration is $a = v/t = 20/10 = 2$ m/s².)

Question 4

A car starts from rest with constant acceleration $a_1 = 3$ m/s² at the same instant a bus passes it at constant $v_2 = 16$ m/s. Setting the positions equal, $\tfrac{3}{2}t_a^2 = 16\,t_a$ , gives the overtaking time. Approximately when does the car overtake the bus?

✅ Correct! Dividing by t_a gives (3/2)t_a = 16, so t_a = 32/3 ≈ 10.7 s.

❌ Not quite. Divide both sides by t_a first: (3/2)t_a = 16 → t_a = 32/3 ≈ 10.7 s. Don't stop at 32 (that's 2×16 before dividing by 3).

Show solution

Solution:

Set the two position functions equal. The car: $x_1(t) = \tfrac{1}{2}(3)t^2 = \tfrac{3}{2}t^2$ . The bus: $x_2(t) = 16t$ . They meet (after $t=0$ ) where

$\tfrac{3}{2}t_a^2 = 16\,t_a$

Divide both sides by $t_a$ (valid since $t_a \neq 0$ ):

$\tfrac{3}{2}t_a = 16 \;\;\Rightarrow\;\; t_a = \frac{2(16)}{3} = \frac{32}{3} \approx 10.7 \text{ s}$

Substituting back, the meeting position is $x_1(t_a) = \tfrac{3}{2}t_a^2 = \tfrac{512}{3} \approx 170$ m.

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