If my memory serves me well, if we want to know the velocity that an object is traveling, we must know the <span>direction and speed. Velocity includes two these points listed in the previous sentence which means the answer is D.</span>
<span>We put a motion detector at </span>one end of the track<span> and put a cart on the track. ... Next, we put a motorized fan on the cart and let it push the cart down the track. ... This is what I would expect based on the velocity graph, since </span>acceleration<span> equals the slope of the velocity graph, which remains</span>constant<span> in time.</span>
Your rotational speed would still be the same. This is because all parts of the Ferris wheel rotate together. Your linear speed however would change. That is a function of radius. But the question is asking about rotational speed and that does not change in this situation
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
