Question 41: Motion in 1D

Question

If $$\mathrm{t}=\sqrt{x}+4$$, then $$\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$$ is :

Accepted Answer

Given,
$t=\sqrt{x}+4$, Squaring on both
$$
\begin{aligned}
& x=(t-4)^2=t^2-8 t+16 \\
& \frac{d x}{d t}=2 t-8\\
& 	ext { at } t=4 \\
& \frac{d x}{d t}=8-8=0
\end{aligned}
$$

Answer

4

Answer

zero

Answer

8

Answer

16

Motion in 1D

If $\mathrm{t}=\sqrt{x}+4, then \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$ is :

Choose the correct answer: