Two students walk in the same direction along a straight path, at a speed-one at 0.90 m/s and the other at 1.90 m/s. Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away?
The faster student arrives at the destination 7 minutes and 36 secs (approximately) sooner than the slower student.
Let the two students be A and B
i) Speed of A = 0.90 m/s ---- Let this be s1
ii) Speed of B is 1.90 m/s -------Let this be s2
iii) Distance to be covered = 780 m -----let this be
We need to find out the time taken by A and B to cover this distance to know how sooner the faster student arrive at destination. Let the time be t1 and t2 respectively.
The equation for speed is
Time taken =
Student A takes more time than student B, i.e. B reaches first.
We find the difference t1 - t2
In minutes ------
i.e. 7 minutes and 36 seconds
Answer : Student B reaches the destination 7 minutes 36 seconds (approximately) sooner than student A.
Note : all the values are truncated up to two decimal places without rounding.