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?

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 / $s p e e d$ so t1 = (d/ s)$i . e . t 1 =$(780/ 0.90)$=$866.66 sec.

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

$t 2 = \left(\frac{d}{s}\right)$ i.e. t2 = $\left(\frac{780}{1.90}\right)$ = $410.52$ 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 ------ $\frac{456.14}{60}$ = $7.60$ 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.