Asked by Genevieve
A particle has r(0)= (4m)j and v(0)= 92 m/s)i
If uts acceleration is constant and given by a=-(2 m/s^2)(i+j), at what time t does the particle first cross the x axis?
If uts acceleration is constant and given by a=-(2 m/s^2)(i+j), at what time t does the particle first cross the x axis?
Answers
Answered by
Damon
r(t) = x(t) i + y(t) j
r(0) = 0 i + 4 j
r' = x' i + y' j
r'(0) = 92 i + 0 j
r'' = x" i + y" j
= -2 i - 2 j always
integrate acceleration
velocity =
r' = [-2 t + x'(0) ] i +[-2 t + y'(0)] j
= [ -2 t +92 ] i - 2 t j
integrate velocity
r = [-t^2 +92 t ] i +[-t^2 +4 ] j
when is -t^2 + 4 = 0?
when t = 2, the end
r(0) = 0 i + 4 j
r' = x' i + y' j
r'(0) = 92 i + 0 j
r'' = x" i + y" j
= -2 i - 2 j always
integrate acceleration
velocity =
r' = [-2 t + x'(0) ] i +[-2 t + y'(0)] j
= [ -2 t +92 ] i - 2 t j
integrate velocity
r = [-t^2 +92 t ] i +[-t^2 +4 ] j
when is -t^2 + 4 = 0?
when t = 2, the end
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