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: Option d.

Step-by-step 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=(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+(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 (x_1)/(x_2)=(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=(v_r-v_p)/(t_1+t_2)\ \ \ .........[3]`

We now write the equation for both distances .

`\displaystyle x_1=v_pt_1+(at_1^2)/(2)`

`\displaystyle x_2=v_qt_2+(at_2^2)/(2)`

Using [1] again

`\displaystyle x_2=(v_p+at_1)t_2+(at_2^2)/(2)`

Since

`x_2=2x_1`

We have

`\displaystyle (v_p+at_1)t_2+(at_2^2)/(2)=2\left (v_pt_1+(at_1^2)/(2)\right )`

Operating

`\displaystyle v_pt_2+at_1t_2+(at_2^2)/(2)=2v_pt_1+at_1^2`

Rearranging

`\displaystyle v_pt_2-2v_pt_1=at_1^2-at_1t_2-(at_2^2)/(2)`

Factoring both sides

`\displaystyle v_p(t_2-2t_1)=(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)=(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 (t_1)/(t_2)=1`

The option d. is correct