Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.
a.) assuming that they start at the same point ant the same time, how much sooner does the faster arrive at a 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?
a.) Well, since the faster student is moving at 1.90 m/s and the slower student is moving at 0.90 m/s, the faster student has a speed advantage of 1 m/s. So, to find out how much sooner the faster student arrives at a destination 780 m away, we just need to divide the distance (780 m) by the speed advantage (1 m/s).
780 m / 1 m/s = 780 seconds
So, the faster student arrives 780 seconds sooner than the slower student. But let's make it more relatable. This is like the faster student arriving in a flash while the slower student is still contemplating whether they left the stove on!
b.) Now, to figure out how far the students would have to walk for the faster student to arrive 5.50 minutes (or 330 seconds) before the slower student, we'll reverse our previous formula.
Distance = Speed x Time
Since the faster student arrives 330 seconds earlier, we'll make the time for the slower student 330 seconds longer. We can represent that as:
Time for slower student = Time for faster student + 330 seconds
Now, let's say the distance they have to walk is D meters. So, for the slower student, we have:
D = 0.90 m/s x (Time for faster student + 330 seconds)
And for the faster student, we have:
D = 1.90 m/s x Time for faster student
Let's solve these equations simultaneously!
0.90 m/s x (Time for faster student + 330 seconds) = 1.90 m/s x Time for faster student
0.90 m/s x Time for faster student + 297 m = 1.90 m/s x Time for faster student
297 m = 1.00 m/s x Time for faster student
Time for faster student = 297 seconds
Now, we can find the distance using the speed of the faster student:
D = 1.90 m/s x 297 seconds
D = 562.3 m
So, the students would have to walk approximately 562.3 meters for the faster student to arrive 5.50 minutes (or 330 seconds) before the slower student. Who knew math could make you take longer walks?
To solve these problems, we can use the formula:
Time = Distance / Speed
a) To find out how much sooner the faster student arrives at a destination 780m away, we need to calculate the time taken by each student.
For the slower student:
Time = Distance / Speed
Time = 780m / 0.90m/s
Time = 866.67s
For the faster student:
Time = Distance / Speed
Time = 780m / 1.90m/s
Time = 410.53s
The difference in time is:
Time difference = Time taken by slower student - Time taken by faster student
Time difference = 866.67s - 410.53s
Time difference = 456.14s
Therefore, the faster student arrives 456.14 seconds (s) sooner at the destination.
b) To find out how far the students have to walk so that the faster student arrives 5.50 minutes (min) before the slower student, we first need to convert the time into seconds, as the speeds are given in meters per second.
5.50 min = 5.50 min * 60s/min = 330s
Now, let's assume the distance they have to walk is "x."
For the slower student:
Time = Distance / Speed
330s = x / 0.90m/s
330s * 0.90m/s = x
297m = x
For the faster student:
Time = Distance / Speed
330s = x / 1.90m/s
330s * 1.90m/s = x
627m = x
Therefore, the students would have to walk a distance of 297 meters so that the faster student arrives 5.50 minutes before the slower student.
B. 570m
(a) distance = speed x time
or time = distance/speed
The two times can be found by dividing the distance by each speed. Subtract the two times.
(b) Let t = time it takes the slower student.
5.50 min = 330 s
The distance is
0.90t = 1.90(t-330)
Solve for t, then substitute into 0.90t