Physics please help???????

Question

I am very lost with this question can someone please help me.:) and please explain what u do in detail because i have no idea what to do. thanks please take ur time and make sure it is correct thanks

Two students walk in the same direction along a straight path, at a constant speed-one at 0.90m/s and the other at 1.90m/s.

a. Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780m away???

b. How far would the students have to walk so that the faster student arrives 5.50min before the slower student ???

Answer 1

a)
divde 780 by 0.90 >>> 866.666 seconds for student 1
divide 780 by 1.90 >>> 410.526 seconds for student 2

find the difference between the two times
866.66.-410.52>>> approximately 456.14 seconds difference or 7.60 minutes


b)
5.50 min = 330 seconds
let x = distance walked
x/0.9 - x/1.9 = 330
1.11x - .52x = 330
.59x = 330
x = 559.32 meters

Answer 2

distance = speed * time

faster student:
780 = 1.9t
t =~ 410.53s

slower students
780 = 0.9t
t =~ 866.67s

866.67 - 410.53 = 456.14s
so the faster student arrives a destination of 780m about 556.14s soon than the slower students


5.50min = 330s

let x be the distance both of them cover. The faster students arrives a a certain distance 330s soon than the slower. So if you take the time it takes for the faster to cover a distance and subtract the time it takes the slower student to cover a same distance, it'd give you 330s

time (faster student) - tim (slower student) = 330

from the equation distance = speed * time, we have:
time = distance/speed

time of the faster student
t = x/1.9

time of the slower student
t = x/.9

x/.9 - x/1.9 = 330
(100/171)x = 330
x =~ 564.3m

hope this helps

Answer 3

The rates at which the students walk can be expressed as linear equations. Y1 = 0.9X and Y2 = 1.9X where x is in seconds and y is in meters.

You could graph Y1 and Y2 on your graphing calculator. If you also graphed Y3 = 780, you could find the intersection point between Y1 and Y3 (2nd key then "calc" on a TI calculator) and the intersection of Y2 and Y3. You'll find that for walking 780 meters, the faster student gets there 456.2 seconds earlier.

Part b is more difficult.... I spent about 10 min. thinking it over then finally gave up and went for brute force. I opened up an excel spreadsheet and created 3 columns. 1st column was time in seconds, second column was the equation Y1=0.9X; third column was Y2=1.9X. Using the click and drag technique, I populated the columns down to about 1000 seconds. Then I just began a tedious comparison routine.... I selected a time from the first column then checked the distance in the second column - I took that distance and looked it up in the 3rd column and found the time associated with it... I subtracted the times and if they didn't equal 330 seconds (5.5 min.) I adjusted my search. After 9 trys (!) I finally arrived at the fact that they must travel 564.3 meters for one student to arrive 5.5 min ahead of the other.... I KNOW there is an easier way to do this I just need to think about it some more

Answer 4

Okay. Let the first person have a speed v1 = 0.9 m/s and the second person have a speed v2 = 1.9 m/s. Now distance, speed and time are related by:

d = v*t so setting d = 780 m whe get:

t1 = d/v1 = 866.7 seconds and t2 = d/v2 = 410.5 seconds

Then t2 - t1 = 456.1 seconds so student 2 is waiting around for about 7.6 minutes.

Now if student 2 arrives at a point 5.5 minutes before student 1, we can find the distance by usng the above relationship for distance, speed and time:

d = v1 *(t+5.5 min) = v2*t convert minutes to seconds

5.5 min = 330 seconds. We first find the time t:
(v2-v1)t = v1*330sec ---> t = v1/(v2-v1) * 330 sec = 297 sec

Then d = v2*297 = 564.3 m

Answer 5

a) T = 780 m * (1 / 0.90 m/s - 1 / 1.90 m/s)

All your're doing here is calculating the time for each one and subtracting. The time is equal to the distance divided by the velocity. Since the distance of 780 m is the same for both, I moved it outside the parentheses. Solve for T. My answer is T = 456 s, which is about 7 min 36 sec.

b) (5.50 min * 60 s/min) = X * (1 / 0.90 m/s - 1 / 1.90 m/s)

This is the same equation as in (a), except now your unknown is the distance X instead of time T. Solve for X. Also note that I included the conversion of 60 s/min to make sure your units are consistent. My answer is X = 564 m.

Answer 6

a is 456.14 seconds

b is 559 meters