If θ is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed, then

Question

If θ is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed, then (A) θ=90° (B) 0°<θ<90° (C) 90°<θ<180° (D) 0° ≤ θ ≤ 180°

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/90dc7242b421727b75dd8adff757c2ad-.png" /> θ between v Q is 90°<θ<180°

Q. If $θ$ is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed, then

$θ = 9 0^{\circ}$

$0^{\circ} < θ < 9 0^{\circ}$

$9 0^{\circ} < θ < 18 0^{\circ}$

$0^{\circ} \leq θ \leq 18 0^{\circ}$

$θ$ between $v & Q$ is $9 0^{\circ} < θ < 18 0^{\circ}$

Q. If θ is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed, then

θ=90∘

0∘<θ<90∘

90∘<θ<180∘

0∘≤θ≤180∘