Plxxxx answer this question..?...its urgent.......soooonnnn....
Plxxxx answer this question..?...its urgent.......soooonnnn....
Tanisha , 5 Years ago
Grade 11
Mechanics> Plxxxx answer this question..?...its urge...
2 AnswersAditya Gupta
Last Activity: 5 Years ago
a= sin(pit)+cos(pit)
but a=dv/dt
so dv/dt= sin(pit)+cos(pit)
or dv= (sin(pit)+cos(pit))dt
now integrate both sides
v= (sin(pit) – cos(pit))/pi + C
but at t=0, it is at rest or v=0 so that
0 = C – 1/pi or C= 1/pi
so v= (sin(pit) – cos(pit))/pi + 1/pi= (sin(pit) – cos(pit) + 1)/pi
now to find the max value of v, we simply need to find the max value of sin(pit) – cos(pit), which is root(1^2+(-1)^2)= root2
so, v(max)= (root2 + 1)/pi m/s
option 3 is correct
kindly approve :)
aswanth nayak
Last Activity: 5 Years ago
Dear Student
a= sin(pit)+cos(pit)
but a=dv/dt
so dv/dt= sin(pit)+cos(pit)
or dv= (sin(pit)+cos(pit))dt
now integrate both sides
v= (sin(pit) – cos(pit))/pi + C
but at t=0, it is at rest or v=0 so that
0 = C – 1/pi or C= 1/pi
so v= (sin(pit) – cos(pit))/pi + 1/pi= (sin(pit) – cos(pit) + 1)/pi
now to find the max value of v, we simply need to find the max value of sin(pit) – cos(pit), which is root(1^2+(-1)^2)= root2
so, v(max)= (root2 + 1)/pi m/s
Regards
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