Asked by Margie
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.) 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?
Answers
Answered by
GK
(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
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
Answered by
Paris
B. 570m
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