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Grade 11Mechanics

Two cars P and Q start from a point at time in a straight line and their positons are represented by Xp(t)=at+bt² and Xq(t) = ft–t².At what time do the cars have the same velocity?full solution

Profile image of Harshit Raj
9 Years agoGrade 11
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1 Answer

Profile image of Vikas TU
9 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)