8) \( f(t)=-4 \sin 2 t \), and \( v(0)=2, x(0)=-3 \), then \( x(\pi)= \) (a) -3 (b) zero (c) 2 (d) 3 9) If \( \mathrm{v}=3 \mathrm{t}-2 \), then s in the interval \( [0,2] \) is \( \qquad \) unit length. (a) 1 (b) 2 (c) 3 (d) 4 10)If \( v=3 t^{2}-2 t \), then the distance covered within the interval \( [0,2] \) \( \qquad \) unit length. (a) \( \frac{4}{27} \) (b) 4 (c) \( \frac{112}{27} \) (d) \( \frac{116}{27} \) 11) If \( \mathrm{v}=\mathrm{t}^{3}-3 \mathrm{t}^{2}+2 t \), then the distance covered within the time interval \( [0,3] \) \( \qquad \) unit length. (a) \( \frac{1}{4} \) (b) \( \frac{1}{2} \) (c) \( \frac{9}{4} \) (d) \( \frac{11}{4} \) 12)If \( a=3, v_{0}=-1 \), then \( s \) within the time interval \( [0,2] \) \( \qquad \) unit lengh (a) \( \frac{1}{6} \) (b) 4 (C) \( \frac{25}{6} \) (d) \( \frac{13}{3} \) 13)fra=3,\( v_{0}=-1 \), then the distance covered within the time interval \( [0,2] \) unit length. (a) \( \frac{1}{6} \) (b) 4 (C) \( \frac{25}{6} \) (d) \( \frac{13}{3} \)
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Bonus Knowledge
The function \( f(t) = -4 \sin 2t \) represents a simple harmonic motion, where the amplitude is 4, indicating how far it oscillates from its equilibrium position. Knowing that the sine function oscillates between -1 and 1, you can plug in \( t=\pi \) to find \( f(\pi) = -4 \sin(2\pi) = 0 \). Using initial conditions, we find out the position after straightforward calculations, leading to your answer being (a) -3. Integrating velocity functions over an interval gives you the distance traveled. For \( v = 3t - 2 \), we can set up the integral from 0 to 2. After evaluating this integral, you'll get the total distance traveled in that time frame. This can guide you in checking the total motion compared to displacement to calculate s accurately, helping you select the right answer from the options.
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