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Physics-

General

Easy

Question

An object moves at a constant speed along a circular path in a horizontal $X Y$ plane, with the centre at the origin. When the object is at $x equals negative 2 m$ , its velocity is $negative left parenthesis 4 m divided by s right parenthesis stack j with hat on top$ . What is the object’s acceleration when it is $y equals 2 m$

$- (8 m / s^{2}) \hat{j}$
$- (8 m / s^{2}) \hat{i}$
$- (4 m / s^{2}) \hat{j}$
$(4 m / s^{2}) \hat{i}$

The correct answer is: $negative left parenthesis 8 m divided by s to the power of 2 end exponent right parenthesis stack j with hat on top$

The radius of circular path is $2 m$ and the speed of the object is $4 m divided by s$
The magnitude of acceleration is
$a equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 16 over denominator 2 end fraction equals 8 m divided by s to the power of 2 end exponent$
The acceleration is directed towards the centre
Therefore, when an object is at $y equals 2 m$ , its acceleration is $negative 8 stack j with hat on top blank m divided by s to the power of 2 end exponent$

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