This is a two-part question, and I got the first part, but I do not know how to do the second part.

Two students walk in the same direction along a straight path, at a constant speed- one at 0.90 m/s and the other at 1.90 m/s.
a) Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at destination 780 m away?
b) How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

I calculated part A and got 456.2 seconds, but I am unsure how to do part B.

I am wondering what you did on A, if you can't do b?

Let me see your thinking on A.

I divided 780/.90 m/s which is 866.7 and then I divided 780/1.9 m/s which is 410.5 and then subtracted 866.7-410.5 to get 456.2 seconds.

This is a two-part question, and I got the first part, but I do not know how to do the second part.

Two students walk in the same direction along a straight path, at a constant speed- one at 0.90 m/s and the other at 1.90 m/s.
a) Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at destination 780 m away?
b) How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

Distance1=speed1*time1
distance2=speed2*time2
subtract the second equation from the first.
distance1-distance2=speed1*time1 -speed2*time2
but the distances are the same.
0=1.9Time1-.90time2
But time2=time1*5.50min*60sec/min
Put that for time 2 in the equation, and solve for time1, and time2.
Now, you can solve for distance

distance=.90*time2

get your big ..

thanks man.

To find the distance the students have to walk so that the faster student arrives 5.50 min before the slower student, we can use the equation we derived earlier:

0 = 1.9 * Time1 - 0.9 * Time2

Since we know Time2 = Time1 * 5.50 min * 60 sec/min, we can substitute Time2 into the equation:

0 = 1.9 * Time1 - 0.9 * Time1 * 5.50 min * 60 sec/min

Now we can solve for Time1:

0 = Time1 * (1.9 - 0.9 * 5.50 * 60)

Simplifying further:

0 = Time1 * (1.9 - 297)

0 = Time1 * (-295.1)

Since we cannot divide by zero, the only solution is for Time1 to equal zero. This means that there is no distance the students can walk where the faster student arrives 5.50 min before the slower student.

To solve part B of the question, you need to find the distance at which the faster student arrives 5.50 minutes before the slower student.

We already know that the time difference between the two students is 5.50 minutes. Let's convert this time to seconds, as the speeds are given in meters per second.

5.50 minutes * 60 seconds/minute = 330 seconds

Now, we can use the equation distance = speed * time to find the distance for the faster student:

distance = 1.90 m/s * time

We can substitute the time of the faster student, which is time2, into this equation:

distance = 1.90 m/s * time2

And we know that the time difference between the students is:

time2 = time1 + 330 seconds

Substituting this into the equation, we get:

distance = 1.90 m/s * (time1 + 330 seconds)

Now, let's use the equation distance = 0.90 m/s * time1 for the slower student:

distance = 0.90 m/s * time1

We want to find the distance at which the faster student arrives 5.50 minutes before the slower student, so the distances must be the same. Setting the two distance equations equal to each other:

1.90 m/s * (time1 + 330 seconds) = 0.90 m/s * time1

Now, we can solve for time1:

1.90 time1 + 627 - 0.90 time1 = 0

0.90 time1 = -627

time1 = -627 / 0.90

time1 ≈ -696.67 seconds

Notice that this result is negative, which means it is not physically meaningful. It suggests that the faster student cannot arrive 5.50 minutes before the slower student. This might be due to the difference in their speeds, as the faster student will always be catching up to the slower student. Therefore, there is no distance at which the faster student arrives 5.50 minutes before the slower student.