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) 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 0.90tposted by GK
B. 570mposted by Paris