Answer:
Answer: Option d.
Explanation:
Accelerated Motion
When an object changes its spped in the same amounts in the same times, the acceleration is constant and its value is
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
Where [tex]v_f, v_o, t[/tex] are the final speed, initial speed, and time taken to change them, respectively
From the above equation we can know
[tex]v_f=v_o+at[/tex]
The distance traveled is computed as
[tex]\displaystyle x=v_ot+\frac{at^2}{2}[/tex]
The question talks about a car moving in a straight line with constant acceleration. It goes through the points p,q,r such as
[tex]v_p=5\ m/s, v_r=25\ m/s[/tex]
The ratio of the distances traveled in each segment is
[tex]\displaystyle \frac{x_1}{x_2}=\frac{1}{2}[/tex]
being [tex]x_1[/tex] the distance from p to q and [tex]x_2[/tex] the distance from q to r
It means that
[tex]x_2=2x_1[/tex]
From the equation for speed
[tex]v_q=v_p+at_1\ \ \ ..........[1][/tex]
[tex]v_r=v_q+at_2\ \ \ ..........[2][/tex]
Replacing [1] into [2]
[tex]v_r=v_p+at_1+at_2[/tex]
[tex]v_r=v_p+a(t_1+t_2)[/tex]
Solving for a
[tex]\displaystyle a=\frac{v_r-v_p}{t_1+t_2}\ \ \ .........[3][/tex]
We now write the equation for both distances .
[tex]\displaystyle x_1=v_pt_1+\frac{at_1^2}{2}[/tex]
[tex]\displaystyle x_2=v_qt_2+\frac{at_2^2}{2}[/tex]
Using [1] again
[tex]\displaystyle x_2=(v_p+at_1)t_2+\frac{at_2^2}{2}[/tex]
Since
[tex]x_2=2x_1[/tex]
We have
[tex]\displaystyle (v_p+at_1)t_2+\frac{at_2^2}{2}=2\left (v_pt_1+\frac{at_1^2}{2}\right )[/tex]
Operating
[tex]\displaystyle v_pt_2+at_1t_2+\frac{at_2^2}{2}=2v_pt_1+at_1^2[/tex]
Rearranging
[tex]\displaystyle v_pt_2-2v_pt_1=at_1^2-at_1t_2-\frac{at_2^2}{2}[/tex]
Factoring both sides
[tex]\displaystyle v_p(t_2-2t_1)=\frac{a}{2}\left (2t_1^2-2t_1t_2-t_2^2 \right )[/tex]
Replacing the equation [3] for a :
[tex]\displaystyle 2v_p(t_2-2t_1)=\frac{v_r-v_p}{t_1+t_2}\left (2t_1^2-2t_1t_2-t_2^2 \right )[/tex]
Replacing [tex]v_p=5,\ v_r=25,\ v_r-v_p=20[/tex], and operating the denominator
[tex]\displaystyle 10(t_2-2t_1)\left (t_1+t_2 \right )=20\left (2t_1^2-2t_1t_2-t_2^2 \right )[/tex]
Operating and simplifying, we get a second-degree equation
[tex]\displaystyle t_2^2+t_1t_2-2t_1^2=0[/tex]
Factoring
[tex](t_2-t_1)(t_2+2t_1)=0[/tex]
The only positive and valid answer is
[tex]t_2=t_1[/tex]
Or equivalently
[tex]\displaystyle \frac{t_1}{t_2}=1[/tex]
The option d. is correct
