One ship, sailing east with a speed of 7.5 m/s, passes a certain point at 8A.Mand a second ship sailing north at the same speed, passed the same point at 9:30 A.M. At What time are they closest together and what is the distance between them then?

let a be the distance from a to the ship1

and b be the distance from a to the ship2
now c is the distance between a and b.
c^2=a^2+b^2
let time t be the time after 8AM, so t=0 is 8 am, 9:30 is 1.5
I am going to use speed 7, you can do it yourself with speed 7.5
c^2=(7*t )^2 + (7(t-1.5))^2
c^2=49(t^2+t^2-3t+2.25)
taking the derivative and setting dc/dt=0, then
0=49(4t-3)
time t=3/4 hr or 45 min.
and at that time,
c=7*sqrt(18/16-9/4+2.25)=7.42
check this.

They are closest at 8:45 am and the distance between them is 28637.82m

JEE mains ka bhi dijyena sir

To figure out the time when the two ships are closest together and the distance between them, we need to analyze their positions based on their speeds and the times they pass the certain point.

Let's start with the first ship, which is sailing east at a speed of 7.5 m/s and passes the point at 8:00 A.M. We can assume the position of this ship at any given time as x = 7.5t, where x is the distance traveled by the ship in meters and t is the time elapsed since 8:00 A.M. in seconds.

Next, we have the second ship sailing north also at a speed of 7.5 m/s, but it passes the same point at 9:30 A.M. We can assume the position of this ship at any given time as y = 7.5(t - 5400), where y is the distance traveled by the ship in meters and t is the time elapsed since 8:00 A.M. in seconds (converted to 9:30 A.M.).

Now, to calculate when the two ships are closest together, we need to find the time when their positions, x and y, are the closest. This happens when the rate of change of the distance between them is zero.

To find this time, we have to derive the expression for the distance between the two ships as a function of time, then set its derivative equal to zero.

The distance between the two ships at any given time is given by:
d = √(x^2 + y^2)

Taking the derivative of d with respect to time t, we have:
ddt = (1/2) * (2x * dxdt + 2y * dydt)

Setting ddt equal to zero and solving for t, we can find the time when the ships are closest together.

Once we have the time, we can substitute it back into the expressions for x and y to calculate the distance between the two ships at that time.

Remember to convert the time to hours and minutes to determine the exact time when the ships are closest together.

They're both going 7.5 miles per second but 1 hour and 30 minutes apart. So the closest time together is 12:00 PM because at 9:30 + 2:50, because that is how far apart they are, which equals 12:00 PM.

Here:

Closest at 12:00 PM

Distance is 2.5 m/s