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.

Related questions
Impact of this question
22238 views around the world
You can reuse this answer
Creative Commons License