Question

A car move with uniform acceleration along a straight line pqr .Its speed at p and r are 5m/s and 25m/s respectively if pq:qr is 1:2 the ratio of the time taken to travel pq and qr is

Answer 1

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  

$\displaystyle a=\frac{v_f-v_o}{t}$

Where $v_f, v_o, t$ are the final speed, initial speed, and time taken to change them, respectively

From the above equation we can know

$v_f=v_o+at$

The distance traveled is computed as

$\displaystyle x=v_ot+\frac{at^2}{2}$

The question talks about a car moving in a straight line with constant acceleration. It goes through the points p,q,r such as

$v_p=5\ m/s, v_r=25\ m/s$

The ratio of the distances traveled in each segment is

$\displaystyle \frac{x_1}{x_2}=\frac{1}{2}$

being $x_1$ the distance from p to q and $x_2$ the distance from q to r

It means that

$x_2=2x_1$

From the equation for speed

$v_q=v_p+at_1\ \ \ ..........[1]$

$v_r=v_q+at_2\ \ \ ..........[2]$

Replacing [1] into [2]

$v_r=v_p+at_1+at_2$

$v_r=v_p+a(t_1+t_2)$

Solving for a

$\displaystyle a=\frac{v_r-v_p}{t_1+t_2}\ \ \ .........[3]$

We now write the equation for both distances .

$\displaystyle x_1=v_pt_1+\frac{at_1^2}{2}$

$\displaystyle x_2=v_qt_2+\frac{at_2^2}{2}$

Using [1] again

$\displaystyle x_2=(v_p+at_1)t_2+\frac{at_2^2}{2}$

Since

$x_2=2x_1$

We have

$\displaystyle (v_p+at_1)t_2+\frac{at_2^2}{2}=2\left (v_pt_1+\frac{at_1^2}{2}\right )$

Operating

$\displaystyle v_pt_2+at_1t_2+\frac{at_2^2}{2}=2v_pt_1+at_1^2$

Rearranging

$\displaystyle v_pt_2-2v_pt_1=at_1^2-at_1t_2-\frac{at_2^2}{2}$

Factoring both sides

$\displaystyle v_p(t_2-2t_1)=\frac{a}{2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Replacing the equation [3] for a :

$\displaystyle 2v_p(t_2-2t_1)=\frac{v_r-v_p}{t_1+t_2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Replacing $v_p=5,\ v_r=25,\ v_r-v_p=20$, and operating the denominator

$\displaystyle 10(t_2-2t_1)\left (t_1+t_2 \right )=20\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Operating and simplifying, we get a second-degree equation

$\displaystyle t_2^2+t_1t_2-2t_1^2=0$

Factoring

$(t_2-t_1)(t_2+2t_1)=0$

The only positive and valid answer is

$t_2=t_1$

Or equivalently

$\displaystyle \frac{t_1}{t_2}=1$

The option d. is correct