A body takes the following path and moves with constant speed. If $a_{A}$ and $a_{B}$ are the magnitude of its radial acceleration at A and B, then.

a_{A} = a_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a_{A} < a_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a_{A} > a_{B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $a_{A} > a_{B}$
The radius of curvature is the radius of the osculating circle at that point of the curve.

We can clearly see that $R_{A} < R_{B}$
$∴ \frac{v^{2}}{R_{A}} > \frac{v^{2}}{R_{B}}$ (as v is the speed and is constant)
$\Rightarrow a_{A} > a_{B}$

Suggest Corrections

Similar questions

A body takes the following path and moves with constant speed. If $a_{A}$ and $a_{B}$ are the magnitude of its radial acceleration at A and B, then.

A body takes the following path and moves with constant speed. If $a_{A}$ and $a_{B}$ are the magnitude of its radial acceleration at A and B.

Join BYJU'S Learning Program