Solve the problem.
-The position of a particle moving along a coordinate line is $s = \sqrt { 2 + 2 t }$, with $s$ in meters and $t$ in seconds. Find the particle's velocity at $t = 1 \mathrm { sec }$.
A) $\frac { 1 } { 2 } \mathrm {~m} / \mathrm { sec }$
B) $1 \mathrm {~m} / \mathrm { sec }$
C) $- \frac { 1 } { 2 } \mathrm {~m} / \mathrm { sec }$
D) $\frac { 1 } { 4 } \mathrm {~m} / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The velocity of the particle is the derivative of its position with respect to time, $v = \frac{ds}{dt}$. Differentiating $s = \sqrt{2 + 2t}$ with respect to $t$ gives $v = \frac{1}{2\sqrt{2 + 2t}} \cdot 2 = \frac{1}{\sqrt{2 + 2t}}$. Substituting $t = 1$ sec into this expression yields $v = \frac{1}{\sqrt{2 + 2(1)}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$ m/sec.

Solve the Problem s=2+2ts = \sqrt { 2 + 2 t }s=2+2t​

Solve the Problem $s = \sqrt { 2 + 2 t }$