Joseph Krupa

2021-12-14

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What are (a) its speed at the first point and (b) its acceleration?

limacarp4

a) Since that is a motion with a constant acceleration, we can use the following two equations
${v}_{x}={v}_{x,0}+{a}_{x}t$
$x-{x}_{o}={v}_{x,o}t+\frac{1}{2}{a}_{x}{t}^{2}$
where () is the speed at the second point, $\left(x-{x}_{o}=70\right)$ is the traveled distance and t=6 s. Substitutte for ${v}_{x},{a}_{x}$ and t into the two equations to get the following
$15\frac{m}{s}={v}_{x,o}+\left(6s\right){a}_{x}$
rearrange the equation to isolate ${a}_{x}$

Substitute for ${a}_{x}$ from equation (1) into equation (2)

rearrange the equation and solve to find ${v}_{x,o}$

b) The acceleration can be determined by substituting the value of ${v}_{x,o}$ into equation

karton

(a) The speed of the antelope at the first point is ${v}_{1}\approx -8.34$ m/s.
(b) The acceleration of the antelope is $a\approx 3.89$ m/s².
Explanation:
$d$ = distance between the two points ($d=70.0$ m)
$t$ = time taken to cover the distance ($t=6.00$ s)
${v}_{2}$ = speed at the second point (${v}_{2}=15.0$ m/s)
(a) the speed at the first point (${v}_{1}$)
(b) the acceleration ($a$)
We can start by using the formula for displacement with constant acceleration:
$d={v}_{1}t+\frac{1}{2}a{t}^{2}$
Since the antelope starts from rest at the first point, its initial speed (${v}_{1}$) is 0. Plugging in the given values, we have:
$70.0=0·6.00+\frac{1}{2}a·\left(6.00{\right)}^{2}$
Simplifying the equation, we get:
$70.0=18.0a$
To find $a$, we divide both sides of the equation by 18.0:
$\frac{70.0}{18.0}=a$
Hence, the acceleration of the antelope is:
$a\approx 3.89\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
For part (a), we can use the formula for velocity with constant acceleration:
${v}_{2}={v}_{1}+at$
We already know ${v}_{2}$ and $t$, and we just found $a$. Substituting these values, we can solve for ${v}_{1}$:
$15.0={v}_{1}+3.89·6.00$
Simplifying the equation:
$15.0={v}_{1}+23.34$
To isolate ${v}_{1}$, we subtract 23.34 from both sides:
${v}_{1}=15.0-23.34$
Hence, the speed of the antelope at the first point is:
${v}_{1}\approx -8.34\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Note: The negative sign indicates that the antelope is moving in the opposite direction at the first point.

user_27qwe

Step 1: (a) The speed of the antelope at the first point (${v}_{1}$) can be calculated using the formula for constant acceleration:
${v}_{2}={v}_{1}+at$ where ${v}_{2}$ is the speed at the second point, $a$ is the acceleration, and $t$ is the time taken. Rearranging the equation, we have:
${v}_{1}={v}_{2}-at$
Substituting the given values, we get:
${v}_{1}=15.0\phantom{\rule{0.167em}{0ex}}\text{m/s}-a×6.00\phantom{\rule{0.167em}{0ex}}\text{s}$
Step 2: (b) The acceleration ($a$) can be determined using the formula:
$s=ut+\frac{1}{2}a{t}^{2}$ where $s$ is the distance covered, $u$ is the initial velocity, $t$ is the time, and $a$ is the acceleration. Rearranging the equation, we have:
$a=\frac{2\left(s-ut\right)}{{t}^{2}}$
Substituting the given values, we get:
$a=\frac{2\left(70.0\phantom{\rule{0.167em}{0ex}}\text{m}-{v}_{1}×6.00\phantom{\rule{0.167em}{0ex}}\text{s}\right)}{\left(6.00\phantom{\rule{0.167em}{0ex}}\text{s}{\right)}^{2}}$

alenahelenash

To solve the problem, let's use the following variables:
- $d$ for the distance between the two points (given as 70.0 m).
- $t$ for the time taken to cover the distance (given as 6.00 s).
- ${v}_{2}$ for the speed of the antelope at the second point (given as 15.0 m/s).
We need to find:
(a) The speed of the antelope at the first point (${v}_{1}$).
(b) The acceleration of the antelope ($a$).
(a) To find the speed at the first point, we can use the equation for average speed:
${v}_{\text{avg}}=\frac{d}{t}$
The average speed is the total distance divided by the total time. Since the antelope is moving with constant acceleration, the average speed is equal to the average of the initial and final speeds. So we have:
${v}_{\text{avg}}=\frac{{v}_{1}+{v}_{2}}{2}$
Substituting the given values, we get:
$\frac{70.0}{6.00}=\frac{{v}_{1}+15.0}{2}$
To find ${v}_{1}$, we can rearrange the equation:
${v}_{1}=2\left(\frac{70.0}{6.00}\right)-15.0$
(b) To find the acceleration, we can use the equation:
${v}_{2}={v}_{1}+a·t$
Rearranging the equation to solve for $a$, we have:
$a=\frac{{v}_{2}-{v}_{1}}{t}$
Substituting the values we have found, we get:
$a=\frac{15.0-{v}_{1}}{6.00}$
Now, let's calculate the values of ${v}_{1}$ and $a$.
(a) Calculating ${v}_{1}$:
${v}_{1}=2\left(\frac{70.0}{6.00}\right)-15.0$
(b) Calculating $a$:
$a=\frac{15.0-{v}_{1}}{6.00}$

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