Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is

The correct option is 86400.

First, let's list the total values given in the problem:

Total distance traveled = $224\text{ km}=224000\text{ m}$

Total time taken = $3\text{ hours}=180\text{ minutes}$

The problem states that the travel times for the four parts follow an arithmetic progression (A.P.). Let the times be $t_1$, $t_2$, $t_3$, and $t_4$. The time for the first part is given as $t_1=30\text{ minutes}$. Let the common difference for the time A.P. be $d_t$.

We can write the sum of the times as:

$t_1+(t_1+d_t)+(t_1+2d_t)+(t_1+3d_t)=180$

Substituting $t_1=30$:

$30+(30+d_t)+(30+2d_t)+(30+3d_t)=180$

$120+6d_t=180$

$6d_t=60\Rightarrow d_t=10$

So, the times for the four parts are 30, 40, 50, and 60 minutes.

Similarly, the speeds follow an arithmetic progression. Let the speeds be $v_1$, $v_2$, $v_3$, and $v_4$. The speed for the first part is given as $v_1=960\text{ m/min}$. Let the common difference for the speed A.P. be $d_v$.

The speeds are: $960$, $960+d_v$, $960+2d_v$, and $960+3d_v$ m/min.

The total distance is the sum of the distances traveled in each part (Distance = Speed × Time):

$v_1{t}_1+v_2{t}_2+v_3{t}_3+v_4{t}_4=224000$

$(960\times 30)+(960+d_v)\times 40+(960+2d_v)\times 50+(960+3d_v)\times 60=224000$

$28800+(38400+40d_v)+(48000+100d_v)+(57600+180d_v)=224000$

$172800+320d_v=224000$

$320d_v=224000-172800=51200$

$d_v=\frac{51200}{320}=160$

We need to find the distance traveled in the fourth part, which is $v_4\times t_4$. First, let's determine $v_4$:

$v_4=960+3d_v=960+3\left(160\right)=960+480=1440\text{ m/min}$

The time for the fourth part is $t_4=60\text{ minutes}$. Thus, the distance for the fourth part is:

$d_4=v_4\times t_4=1440\times 60=86400\text{ meters}$

Therefore, the distance she travels in the fourth part is 86400 meters.