## Related questions with answers

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

Solution

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

$v = \sqrt{ v_1^2 +v_2^2 } = c$

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

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

Differentiating gives the acceleration vector

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

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

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

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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