Trains a and b are traveling in the same direction on parallel tracks. train a is traveling at 40 miles per hour and train b is traveling at 48 miles per hour. train a passes a station at 1: 15am. if train b passes the same station at 1: 30am, at what time will train b catch up to train a?

Train B will catch up to Train A at 2:45am

Step-by-step explanation:

Let's build the equations of motion for both train A and train B, starting at 1:30 AM.

These equations have the following format:

In which S is the position after t minutes, S(0) is the initial position and v is the velocity, in miles per hour.

Train A

Traveling at 40 mph, so .

At 1:30, 15 minutes have passed. 15 minutes is 0.25 hours. So

The equation of motion for train A is

Train B

Starts at the position 0.

48 miles per hour, so

The equation of motion for train B is

If train B passes the same station at 1:30am, at what time will train B catch up to train A?

We first have to find how long it takes for train B to catch up to train A.

This is when

It is going to take 1.25 hours for the train to catch up. 1.25 hours is 60 + 0.25*60 = 75 minutes.

So train B will catch up train A 75 minutes after 1:30 am, so at 2:45am

We know that
volume of a cone=(1/3)*area of the base*height
Let
A> area of the base
h---> height of the cone
V=(1/3)*A*h> h=3*V/A
V=6 in³
so
original cone
6=(1/3)*A*h> h=18/A

Find the height of a cone with the same base diameter but a volume of only 3 cubic inches

h=3*V/A> h=3*3/A---> h=9/A
so
the height of the cone is half the original height

the answer is
the height of the cone is half the original height