if the velocity of a particle is v=At+Bt^2 where A & B are constants,then the distance traveled by it between 1sec and 2s is
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here v can be written as dx/dt then
dx/dt = at+bt^2
dx=(at+bt^2)dt
Integrate from 1 to 2 sec.
x= a(2)^2÷2 +b(2)^3÷3 - [a(1)^2÷2 + b(1)^3÷3]
x= 2a+ 8b/3 - a/2 -b/3
x= 3a/2 + 7b/3
dx/dt = at+bt^2
dx=(at+bt^2)dt
Integrate from 1 to 2 sec.
x= a(2)^2÷2 +b(2)^3÷3 - [a(1)^2÷2 + b(1)^3÷3]
x= 2a+ 8b/3 - a/2 -b/3
x= 3a/2 + 7b/3
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