Velocity and Acceleration — Question 10

Question 10 – Solution

Strategy. Begin with $\mathbf v=v\mathbf T$, differentiate, and convert $d\mathbf T/dt$ through arc length.

Step 1: Differentiate velocity $$\mathbf a=\frac d{dt}(v\mathbf T)=v'\mathbf T+v\frac{d\mathbf T}{dt}.$$ By the chain rule, $$\frac{d\mathbf T}{dt}=\frac{d\mathbf T}{ds}\frac{ds}{dt} =(\kappa\mathbf N)v.$$ Therefore $$\boxed{\mathbf a=v'\mathbf T+\kappa v^2\mathbf N}.$$

Step 2: Tangential component Dot with $\mathbf T=\mathbf v/v$: $$a_T=\mathbf a\cdot\mathbf T =\boxed{\frac{\mathbf v\cdot\mathbf a}{\|\mathbf v\|}}.$$ It also equals $v'$, the rate of change of speed.

Step 3: Normal component Orthogonality gives $$\|\mathbf a\|^2=a_T^2+a_N^2.$$ Thus $$\boxed{a_N=\sqrt{\|\mathbf a\|^2-a_T^2}} =\boxed{\frac{\|\mathbf v\times\mathbf a\|}{\|\mathbf v\|}} =\kappa v^2.$$ The tangential component changes speed; the normal component changes direction. A motion can therefore accelerate even at constant speed if its path bends.