Answer to Question #108999 in Physics for Tumi

Question #108999
Determine the velocity of a point, it’s tangential and normal accelerations in 10 seconds after motion begins if at the moment 3 seconds it’s normal acceleration was 0.9m/s^2. The equation of point motion is S=120+7t+2t^2m
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Expert's answer
2020-04-13T10:06:32-0400
v(t)=ddts=7+2(2t)=7+4tv(t)=\frac{d}{dt}s=7+2(2t)=7+4t

Thus, the velocity of a point in 10 seconds after motion begins


v(10)=7+4(10)=47msv(10)=7+4(10)=47\frac{m}{s}


at(t)=ddtv=4a_t(t)=\frac{d}{dt}v=4

Thus, it’s tangential acceleration in 10 seconds after motion begins


at(10)=4ms2a_t(10)=4\frac{m}{s^2}

an(3)=v(3)2r=(7+4(3))2r=0.9ms2a_n(3)=\frac{v(3)^2}{r}=\frac{(7+4(3))^2}{r}=0.9\frac{m}{s^2}

r=401 mr=401\ m

Thus, it’s normal acceleration in 10 seconds after motion begins


an(10)=(7+4(10))2401=5.5ms2a_n(10)=\frac{(7+4(10))^2}{401}=5.5\frac{m}{s^2}


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