If $\mathrm{t}=\sqrt{x}+4$, then $\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$ is :
Solution
Given,
<br/><br/>$t=\sqrt{x}+4$, Squaring on both
<br/><br/>$$
\begin{aligned}
& x=(t-4)^2=t^2-8 t+16 \\\\
& \frac{d x}{d t}=2 t-8\\\\
& \text { at } t=4 \\\\
& \frac{d x}{d t}=8-8=0
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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