10. The position of an object moving in a straight line is given by x=4t-2t^2 where x is in meters and t in seconds. a) What is the position of the object at t=2 and 4 s? (10 points) b) What is the instantaneous velocity at t=2 (10 points)

a) To find the position of the object at $t=2$ and $t=4$ seconds, we can substitute these values into the equation $x=4t-2t^{2}$.<br /><br />For $t=2$:<br />$x = 4(2) - 2(2)^2 = 8 - 8 = 0$ meters<br /><br />For $t=4$:<br />$x = 4(4) - 2(4)^2 = 16 - 32 = -16$ meters<br /><br />Therefore, the position of the object at $t=2$ seconds is 0 meters, and at $t=4$ seconds is -16 meters.<br /><br />b) To find the instantaneous velocity at $t=2$, we need to find the derivative of the position function $x=4t-2t^{2}$ with respect to time $t$. The derivative of a function gives us the rate of change of the function at a specific point.<br /><br />The derivative of $x=4t-2t^{2}$ is:<br />$\frac{dx}{dt} = 4 - 4t$<br /><br />Now, we can substitute $t=2$ into the derivative to find the instantaneous velocity:<br />$\frac{dx}{dt}\bigg|_{t=2} = 4 - 4(2) = 4 - 8 = -4$ meters per second<br /><br />Therefore, the instantaneous velocity of the object at $t=2$ seconds is -4 meters per second.