centripetal acceleration= w^2 r
tangential accceleration= 1.51E-3*r
magnitude= sqrt((w^2r)^2+(1.51E-3r)^2)
= r*sqrt(w^4+2.28E-6)
= r*E-3*sqrt(4.88+2.28)
= .00267 r =0.50 m/s^2
check all that.
A train is rounding a circular curve whose radius is 1.87E+2 m. At one instant, the train has an angular acceleration of 1.51E-3 rad/s2 and an angular speed of 0.0470 rad/s. Find the magnitude of the total acceleration (centripetal plus tangential) of the train.
2 answers
a(centripetal) = ω^2•R=(0.047)^2•187=0.413 m/s^2.
a(tangential)=ε•R=1.51•10^-3•187=0.282 m/s^2.
a(total) =sqr(a(centripetal)^2+ a(tangential)^2) =0.5 m/s^2.
a(tangential)=ε•R=1.51•10^-3•187=0.282 m/s^2.
a(total) =sqr(a(centripetal)^2+ a(tangential)^2) =0.5 m/s^2.