# Feynman Lectures on Physics: Vol 1, 11-6: acceleration vector

I’m trying to get through 11-6 section of Feynman’s Lectures on Physics, Vol 1, particularly explanation of acceleration vector calculation in his example:

It’s clear that acceleration vector equals Δv/Δt, but calculation of its components (tangent and transverse) is obscure to me.

Maybe I’m missing some elementary math or some insight behind working with limits, but:

1. Why tangent acceleration is dv/dt?
2. How it even possible to multiply a vector by an angle (v•Δθ)?
3. Do I correctly understand that Δθ/Δt gives us an angle velocity and transverse acceleration is calculated as product of linear velocity times angular velocity (v•(Δθ/Δt))?

2. $$v$$ is not a vector, it is the speed. Even if it where a vector, it would just be a multiplication of a vector with some scalar. Remember that an angle in radians is defined by $$\theta = s/r$$, where $$s$$ is the arc length of a circle and $$r$$ the radius. So an angle is just a dimensionless number.
3. Technically yes, but thats not the point. $$\Delta v_\perp = v \Delta \theta$$ is the same equation as in 2, where $$\Delta \theta$$ is the angle, $$\Delta v_\perp$$ the arc length and $$v$$ the radius of a circle. Of course $$\Delta v_\perp$$ is a straight line and not a part of a curved circle, but in the limit of infinitesimally small $$\Delta \theta$$ the difference disappears, because for a very small angle the curved arc of a circle looks like a straight line.
• Thank you, sir. I've found the example to be pretty confusing, as 2 different speeds are used and r1, r2 are used to describe distance from O, but approximated circle with radius R is used to describe the solution. I would decouple images for better clarity. For example, this image describes tangential acceleration separately pretty well and intuitively obvious: bit.ly/Centripetal-Acceleration – oblalex Sep 18 '19 at 21:11