v(t)=dtds=7+2(2t)=7+4t Thus, the velocity of a point in 10 seconds after motion begins
v(10)=7+4(10)=47sm
at(t)=dtdv=4 Thus, it’s tangential acceleration in 10 seconds after motion begins
at(10)=4s2m
an(3)=rv(3)2=r(7+4(3))2=0.9s2m
r=401 m Thus, it’s normal acceleration in 10 seconds after motion begins
an(10)=401(7+4(10))2=5.5s2m
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