JEE Mains · Physics · STD 11 - 2. motion in straight line
Position of an ant ( \(\mathrm{S}\) in metres) moving in \(\mathrm{Y}-\mathrm{Z}\) plane is given by \(S=2 t^2 \hat{j}+5 \hat{k}\) (where \(t\) is in second). The magnitude and direction of velocity of the ant at \(t=1 \mathrm{~s}\) will be :
- A \(16 \mathrm{~m} / \mathrm{s}\) in \(y\)-direction
- B \(4 \mathrm{~m} / \mathrm{s}\) in \(x\)-direction
- C \(9 \mathrm{~m} / \mathrm{s}\) in \(z\)-direction
- D \(4 \mathrm{~m} / \mathrm{s}\) in \(y\)-direction
Answer & Solution
Correct Answer
(D) \(4 \mathrm{~m} / \mathrm{s}\) in \(y\)-direction
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{v}}=\frac{\mathrm{ds}}{\mathrm{dt}}=4 \mathrm{t} \hat{j}\) At \(\mathrm{t}=1 \sec \overrightarrow{\mathrm{v}}=4 \hat{\mathrm{j}}\)
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