Vikas TU
Last Activity: 7 Years ago
Dear Student,In the event that we expect that auto Q begins moving from point Q in the meantime auto P begins from P, at that point the issue turns into somewhat trickier.
(1) for the situation where the two autos are advancing toward each other, the separation gone via auto P is d and the separation auto Q ventures is 120 - d as Mark states.
Presently, remove as far as speed is given for every auto as:
P: d = s1 * 1hr => d = s1
Q: 120 - d = s2 * 1hr => d = - s2 + 120
In this way, s1 = - s2 + 120
Presently, tragically, we have 2 factors (s1, s2) and just 1 condition. In the event that this is all we had, we'd be up the creek without a paddle.
In any case, we have case 2 to help us.
(2) if the autos are moving a similar way, at that point the separation auto P goes in 6 hours is d + 120:
P: d + 120 = s1 * 6hr => d = 6s1 - 120
The separation auto Q voyages is d:
Q: d = s2 * 6hr => d = 6s2
Thus,
6s1 - 120 = 6s2 => s1 - 20 = s2 (separating all terms by 6)
We now have two conditions for our two factors (s1, s2):
s1 = - s2 + 120
s1 - 20 = s2
In the event that we substitute the estimation of s1 in the principal condition for s1 in the second, at that point:
- s2 + 120 - 20 = s2
100 = 2s2
50km/h = s2 (the speed of auto Q)
To discover s1, substitute 50 for s1 in s1 - 20 = s2:
s1 - 20 = 50 => s1 = 70km/hr.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
