Question

Show that if the speed of a particle along a curve is constant, then the velocity and acceleration vectors are orthogonal.

Solution

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Assume some particle moving along an arbitrary path traced out by a vector function r(t)\mathbf r (t). If we assume further that the particle is moving with constant speed then we have

v=v12+v22=cv = \sqrt{ v_1^2 +v_2^2 } = c

Where cc is some constant. That leads to assume a velocity vector to be

v=ccosti+csintj\mathbf v = c \cos t \, \mathbf i + c \sin t \, \mathbf j

Differentiating gives the acceleration vector

a=csinti+ccostj\mathbf a = - c \sin t \, \mathbf i + c \cos t \, \mathbf j

Doing the dot product between the acceleration and velocity vectors, we have

va=(ccosti+csintj)(csinti+ccostj)=c2sintcost+c2sintcost=0\begin{align*} \mathbf v \cdot \mathbf a &= \big ( c \cos t \, \mathbf i + c \sin t \, \mathbf j \big ) \cdot \big ( - c \sin t \, \mathbf i + c \cos t \, \mathbf j \big ) \\\\ &= - c^2 \sin t \cos t + c^2 \sin t \cos t =0 \end{align*}

Since the dot production vanish. The projection of velocity vector onto acceleration vector is zero and the two vectors are orthogonal.

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