Tangential vs Radial Acceleration

A car going around a curve has two perpendicular acceleration components:

• Radial (centripetal) acceleration $a_r = v^{2}/r$, pointing toward the center of the curve.
• Tangential acceleration $a_t = dv/dt$, pointing along the direction of motion (forward if speeding up, backward if slowing down).

The total acceleration is the vector sum of these two perpendicular components, with magnitude:

$|\vec a| = \sqrt{a_r^{2} + a_t^{2}}$

Step-by-Step Solution

Given: $v = 10\;\text{m/s}$, $r = 30\;\text{m}$, $a_t = 2\;\text{m/s}^{2}$ (speeding up).

Find: The radial acceleration, the tangential acceleration, and the magnitude and direction of the total acceleration.

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Step 1 — Compute the radial (centripetal) component.

$a_r = \dfrac{v^{2}}{r} = \dfrac{(10)^{2}}{30} = \dfrac{100}{30} \approx 3.333\;\text{m/s}^{2}$

This is always present whenever the car is turning, even if the speed is constant.

Step 2 — Identify the tangential component.

We're given:

$a_t = 2.000\;\text{m/s}^{2}$

This component would be zero if the car were going around the curve at constant speed.

Step 3 — Compute the magnitude of the total acceleration.

The two components are perpendicular, so combine with the Pythagorean theorem:

$|\vec a| = \sqrt{a_r^{2} + a_t^{2}} = \sqrt{(3.333)^{2} + (2)^{2}}$

$= \sqrt{11.111 + 4.000}$

$= \sqrt{15.111}$

$\approx 3.887\;\text{m/s}^{2}$

Step 4 — Find the direction of the total acceleration.

Measure the angle from the radial direction (the perpendicular toward the center):

$\phi = \arctan\!\left(\dfrac{a_t}{a_r}\right) = \arctan\!\left(\dfrac{2}{3.333}\right) = \arctan(0.6) \approx 30.96°$

So the total acceleration vector tilts about 31° forward (in the direction of motion) away from the pure-radial direction. The car is mostly being pulled toward the center, with a small forward kick from the engine speeding it up.

Step 5 — Express in terms of g's.

$|\vec a| / g = 3.887 / 9.81 \approx 0.396\;g$

About 0.4 g — fairly gentle for a road car. A racing car can handle 2–3 g in turns.

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Answer:

• $a_r = 3.333\;\text{m/s}^{2}$ (centripetal, toward center)
• $a_t = 2.000\;\text{m/s}^{2}$ (forward, along velocity)
• $\boxed{\;|\vec a| = \sqrt{a_r^{2} + a_t^{2}} \approx 3.887\;\text{m/s}^{2}\;}$ at $\approx 30.96°$ ahead of pure-radial

When a car turns and speeds up at the same time, the total acceleration is the vector sum of these two perpendicular components.

Try It

• Adjust the speed, radius, and tangential acceleration sliders.
• Watch the two perpendicular vectors and the resultant total acceleration.
• Set $a_t = 0$ to see pure circular motion (no speed change) — only the radial component remains.
• Try $r$ very large (almost straight line) — the radial component nearly vanishes.