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?

1 Answer
Pallu
Sep 20, 2017

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.

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