MHB How Does Constant Acceleration Affect Antelope Speed Over 70 Meters?

AI Thread Summary
An antelope accelerates uniformly over a distance of 70 meters in 7 seconds, reaching a final speed of 15 m/s. Its initial speed at the starting point is calculated to be 5 m/s. The acceleration is determined to be approximately 1.43 m/s² using the formula a = (v_f - v_0) / t. Various kinematic equations are discussed for solving the problem, confirming the initial speed and acceleration values. The discussion emphasizes the importance of using variables in physics problems before substituting numerical values.
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An antelope moving with constant acceleration covers the
distance between two points $\textbf{70.0m}$ apart in $\textbf{7.00s}$
Its speed as it passes the second point is $\textbf{15.00 m/s}$.
What is its speed at the first point?

The answer to this is 5 m/s

ok this should be real simple but kinda ?

$\displaystyle
\frac{70m}{7s}=\frac{10m}{s}$
so
$\displaystyle\frac{15-5}{2}=10\\
x=5$.

b. What is its acceleration

$\begin{align*}\displaystyle
a&=\frac{v_2-v_1}{t_2-t_1}\\
&=\frac{15-x}{7}\\
&\approx 1.43 m/s
\end{align*}$

kinda got it

suggestions?
 
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kinematics equation for constant acceleration ...

$\Delta x = \bar{v} \cdot t$

$\Delta x = \dfrac{v_0+v_f}{2} \cdot t$

solve for $v_0$, then $a = \dfrac{\Delta v}{t}$
 
ok, I don't think I know what that would look with numbers given?
 
karush said:
ok, I don't think I know what that would look with numbers given?

Sorry, but I don't understand what you mean by that statement ... physics problems are solved using variables, then values are substituted as the last step.

$\Delta x = \dfrac{v_0+v_f}{2} \cdot t \implies v_0 = \dfrac{2\Delta x}{t} - v_f$

now substitute in your given values ... $v_f = 15 \, m/s$, $t = 7 \, s$, and $\Delta x = 70 \, m$

$v_0 = \dfrac{2 \cdot 70 \, m}{7 \, s} - 15 \, m/s = 5 \, m/s$

finally ...

$a = \dfrac{\Delta v}{t} = \dfrac{v_f-v_0}{t} = \dfrac{15 \, m/s - 5 \, m/s}{7 \, s} = \dfrac{10}{7} \, m/s^2$
 
A slightly different way of looking at it. With constant acceleration a, initial speed [math]v_0[/math], initial position [math]x_0[/math], [math]v(t)= at+ v_0[/math] and [math]x(t)= x_0+ v_0t+ (a/2)t^2[/math].

Taking the first point to be [math]x_0= 0[/math], [math]x(7)= 7v_0+ (49/2)a= 70[/math] and [math]v(7)= 7a+ v_0= 15[/math]. Since we only want to determine [math]v_0[/math], multiply the second equation by 7/2 to get the same coefficient of a, (49/2)a+ (7/2)v_0= 105/2, and subtract from the first equation: [math](7- 7/2)v_0= (7/2)v_0= 70- 105/2= 35/2[/math] so [math]v_0= 35/7= 5[/math] m/s.

Put [math]v_0= 5[/math] into either of those equations and solve for a: The simpler is [math]7(5)+ v_0= 35+ v_0= 15[/math] so [math]v_0= 15- 35= -25 m/s^2[/math]. This is negative indicating that the antelope was slowing down.
 
Plot velocity versus time.
The Area under the curve is the distance traveled for a given time interval.

View attachment 7670

Using the formula for volume of a trapezoid...

A =h (a+b)/2

Where:
a=v
b=15 m/s
h=7 s
A=70 m

then

70=7(v+15)/2
140/7=v+15
v=20-15=5
 

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