TS EAMCET · Physics · Laws of Motion
The velocity of an object of mass \(2 \mathrm{~kg}\) is given by \(\mathbf{v}=\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is time in seconds. What will be the direction of net force on the object relative to the positive direction of \(X\)-axis, at the instant when its magnitude is \(20 \mathrm{~N} ?\)
- A \(\tan ^{-1}\left(\frac{1}{2}\right)\)
- B \(\tan ^{-1}\left(\frac{2}{3}\right)\)
- C \(\tan ^{-1}\left(\frac{4}{5}\right)\)
- D \(\tan ^{-1}\left(\frac{3}{4}\right)\)
Answer & Solution
Correct Answer
(D) \(\tan ^{-1}\left(\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
Given, mass of object, \(m_2=2 \mathrm{~kg}\) Velocity, \(\mathbf{v}=\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}\) \(\therefore\) Acceleration of object is given as…
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