Position, Velocity, and Acceleration as Successive Derivatives

The relationship between position, velocity, and acceleration is the most physically intuitive use of derivatives. Velocity is the derivative of position, and acceleration is the derivative of velocity. So if you have a position formula, you can mechanically read off the other two.

The Physics

If position is $s(t)$, then:

$v(t) = \dfrac{ds}{dt} \qquad a(t) = \dfrac{dv}{dt} = \dfrac{d^{2}s}{dt^{2}}$

For the polynomial $s(t) = 5t^{2} + 2t$, we use the power rule: $\dfrac{d}{dt}\bigl[t^{n}\bigr] = n\,t^{n-1}$, plus the linearity of derivatives.

Step-by-Step Solution

Given: $s(t) = 5t^{2} + 2t$.

Find: $v(t)$, $a(t)$, and the values at $t = 0, 1, 2, 3$.

---

Step 1 — Differentiate $s(t)$ term by term to get velocity.

$\dfrac{d}{dt}\bigl[5t^{2}\bigr] = 5 \cdot 2 t^{2-1} = 10\,t$

$\dfrac{d}{dt}\bigl[2t\bigr] = 2$

So:

$v(t) = 10\,t + 2$

Step 2 — Differentiate $v(t)$ to get acceleration.

$\dfrac{d}{dt}\bigl[10\,t\bigr] = 10$

$\dfrac{d}{dt}\bigl[2\bigr] = 0$

So:

$a(t) = 10$

The acceleration is constant — this is uniform acceleration, like a car flooring the gas pedal.

Step 3 — Tabulate values at a few times.

• At $t = 0$: $s = 5(0)^{2} + 2(0) = 0\;\text{m}$, $v = 10(0) + 2 = 2\;\text{m/s}$, $a = 10\;\text{m/s}^{2}$
• At $t = 1$: $s = 5(1) + 2(1) = 7\;\text{m}$, $v = 10 + 2 = 12\;\text{m/s}$, $a = 10\;\text{m/s}^{2}$
• At $t = 2$: $s = 5(4) + 2(2) = 24\;\text{m}$, $v = 10(2) + 2 = 22\;\text{m/s}$, $a = 10\;\text{m/s}^{2}$
• At $t = 3$: $s = 5(9) + 2(3) = 51\;\text{m}$, $v = 10(3) + 2 = 32\;\text{m/s}$, $a = 10\;\text{m/s}^{2}$

The position grows as a parabola (second-power), the velocity as a straight line (first-power), and the acceleration is flat (constant).

Step 4 — Sanity check: integrate $a$ back up to $s$.

If you start with $a = 10$ and integrate twice with the right initial conditions ($v(0) = 2$, $s(0) = 0$), you should recover $s(t) = 5t^{2} + 2t$:

$\int 10\,dt = 10\,t + C_1, \;\;\text{with}\;\;C_1 = 2 \;\Longrightarrow\; v(t) = 10\,t + 2$

$\int (10\,t + 2)\,dt = 5\,t^{2} + 2\,t + C_2, \;\;\text{with}\;\;C_2 = 0 \;\Longrightarrow\; s(t) = 5\,t^{2} + 2\,t \;\;\checkmark$

---

Answer:

$\boxed{\;v(t) = 10\,t + 2 \quad\text{m/s}\;}$

$\boxed{\;a(t) = 10 \quad\text{m/s}^{2}\;}$

The car has a constant acceleration of $10\;\text{m/s}^{2}$, with an initial velocity of $2\;\text{m/s}$ at $t = 0$. Its velocity grows linearly with time, and its position grows as a parabola.

Try It

• Slide the time widget — watch all three quantities update in lockstep on the three stacked graphs.
• The car icon at the top shows the actual physical position moving rightward.
• Notice the slope of the position curve (cyan) at any instant equals the value on the velocity curve (pink) directly below — that's the geometric meaning of the derivative.
• Likewise, the slope of velocity is constant (10), so the acceleration curve (yellow) is a flat line.