Question

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?

Answer 1

The faster student arrives at the destination 7 minutes and 36 secs (approximately) sooner than the slower student.

Explanation:

Let the two students be A and B

Given that
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 `d`
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

Speed = #`(distance travelled` / `time taken)`#

Therefore
Time taken = #`distance travelled` / `speed` # so #t1 = (d/ s)`i.e. t1 =`(780/ 0.90)`=`866.66# sec.

`866.66` sec. is the time taken by student A and

`t2 = (d/ s)` i.e. t2 = `(780/ 1.90)` = `410.52` sec.

`410.52` sec.is the time taken by student B

Student A takes more time than student B, i.e. B reaches first.

We find the difference t1 - t2

`866.66 - 410.52 =456.14` seconds

In minutes ------ `456.14 / 60` = `7.60` 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.