Answer each question appropriately.
-Find the standard equation for the position $s$ of a body moving with a constant acceleration a along a coordinate The following properties are known:
i. $\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } } = \mathrm { a }$,
ii. $\frac { \mathrm { ds } } { \mathrm { dt } } = \mathrm { v } _ { 0 }$ when $\mathrm { t } = 0$, and
iii. $s = s 0$ when $t = 0$,
where $t$ is time, $s _ { 0 }$ is the initial position, and $v _ { 0 }$ is the initial velocity.
A) $s = \frac { a t ^ { 2 } } { 2 } - v _ { 0 } t - s 0$
B) $s = \frac { a t ^ { 2 } } { 2 } + v _ { 0 } t + s _ { 0 }$
C) $s = a t ^ { 2 } + v _ { 0 } t + s 0$
D) $s = \frac { a t ^ { 2 } } { 2 } + s 0$

Question

Quizplus · Accepted Answer

The standard equation for the position $s$ of a body moving with a constant acceleration a along a coordinate is given by:

$$s = \frac {1}{2} at^2 + v_0t + s_0$$

Using the given properties, we have:

i. $\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } } = \mathrm { a }$, which implies $a = \frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } ^ 2 }$

ii. $\frac { \mathrm { ds } } { \mathrm { dt } } = \mathrm { v } _ { 0 }$ when $\mathrm { t } = 0$, which implies $v_0 = \frac { \mathrm { ds } } { \mathrm { dt } }$ at $t = 0$

iii. $s = s 0$ when $t = 0$, which implies that the body starts at an initial position of $s_0$.

Substituting these values into the standard equation, we get:

$$s = \frac {1}{2} (\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } ^ 2 })t^2 + (\frac { \mathrm { ds } } { \mathrm { dt } }\rvert_{t=0})t + s_0$$

$$s = \frac {1}{2} at^2 + v_0t + s_0$$

Therefore, the best choice is B.

Answer Each Question Appropriately sss Of a Body Moving with a Constant Acceleration a Along

Answer Each Question Appropriately $s$ Of a Body Moving with a Constant Acceleration a Along