The position of a particle moving along the x axis is given in centimeters by x=9.75+1.50t^3 , where t is in seconds. Calculate (a) the average velocity during the time interval t=2.00s to t= 3.00 s; (b) the instantaneous velocity at t=2.00s (c) the instanta- neous velocity at t=3.00s (d) the instantaneous velocity at t= 2.50 s: and (e) the instantaneous velocity when the particle is mid- way between its positions at t=2.00s and t=3.00s (f) Graph x versus t and indicate your answers graphically.

To solve this problem, we need to find the average velocity, instantaneous velocities at different times, and the instantaneous velocity when the particle is midway between its positions at $t=2.00s$ and $t=3.00s$. We will also graph $x$ versus $t$ and indicate our answers graphically.<br /><br />(a) The average velocity during the time interval $t=2.00s$ to $t=3.00s$ can be calculated using the formula:<br /><br />$\text{Average Velocity} = \frac{\text{Change in Position}}{\text{Change in Time}}$<br /><br />First, we need to find the positions at $t=2.00s$ and $t=3.00s$:<br /><br />At $t=2.00s$:<br />$x_1 = 9.75 + 1.50(2.00)^3 = 9.75 + 1.50(8.00) = 9.75 + 12.00 = 21.75 \text{ cm}$<br /><br />At $t=3.00s$:<br />$x_2 = 9.75 + 1.50(3.00)^3 = 9.75 + 1.50(27.00) = 9.75 + 40.50 = 50.25 \text{ cm}$<br /><br />Now, we can calculate the average velocity:<br /><br />$\text{Average Velocity} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{50.25 - 21.75}{3.00 - 2.00} = \frac{28.50}{1.00} = 28.50 \text{ cm/s}$<br /><br />(b) The instantaneous velocity at $t=2.00s$ can be found by taking the derivative of the position function with respect to time:<br /><br />$v(t) = \frac{dx}{dt} = 4.50t^2$<br /><br />At $t=2.00s$:<br />$v(2.00) = 4.50(2.00)^2 = 4.50(4.00) = 18.00 \text{ cm/s}$<br /><br />(c) The instantaneous velocity at $t=3.00s$ can be found using the same derivative:<br /><br />$v(t) = 4.50t^2$<br /><br />At $t=3.00s$:<br />$v(3.00) = 4.50(3.00)^2 = 4.50(9.00) = 40.50 \text{ cm/s}$<br /><br />(d) The instantaneous velocity at $t=2.50s$ can be found using the same derivative:<br /><br />$v(t) = 4.50t^2$<br /><br />At $t=2.50s$:<br />$v(2.50) = 4.50(2.50)^2 = 4.50(6.25) = 28.13 \text{ cm/s}$<br /><br />(e) To find the instantaneous velocity when the particle is midway between its positions at $t=2.00s$ and $t=3.00s$, we first need to find the midpoint position:<br /><br />Midpoint position:<br />$x_{\text{mid}} = \frac{x_1 + x_2}{2} = \frac{21.75 + 50.25}{2} = 36.00 \text{ cm}$<br /><br />Now, we need to find the time at which the particle is at this midpoint position. We can do this by setting the position function equal to the midpoint position and solving for $t$:<br /><br />$x = 9.75 + 1.50t^3 = 36.00$<br /><br />Solving for $t$:<br />$1.50t^3 = 36.00 - 9.75$<br />$1.50t^3 = 26.25$<br />$t^3 = \frac{26.25}{1.50}$<br />$t^3 = 17.50$<br />$t = \sqrt[3]{17.50} \approx 2.65 \text{ s}$<br /><br />Now, we can find the instantaneous velocity at this time:<br /><br />$v(2.65) = 4.50(2.65)^2 = 4.50(7.0225) = 31.60 \text{ cm/s}$<br /><br />(f) To graph $x$ versus $t$, we can plot the position function $x = 9.75 + 1.50t^3$ on a coordinate plane. The graph will show the position of the particle at different times. We can then indicate the answers graphically by marking the positions at $t=2.00s$, $t=3.00s$, and the midpoint position.