Derivatives And 1d Motion

To find the time when the cannonball reaches its maximum height, we need to find the derivative of the height function and set it equal to zero. This is because the maximum height occurs when the velocity of the cannonball is zero. Therefore, the equation we need to solve is h'(t) = 0.

The maximum vertical velocity of the cannonball can be determined by finding the derivative of the height function with respect to time and setting it equal to zero. Taking the derivative of h(t) = 10 + 207t - 4.9t^2 gives us v(t) = 207 - 9.8t. Setting v(t) = 0 and solving for t, we find t = 21.12 seconds. Substituting this value back into the derivative, we find v(21.12) = 207 - 9.8(21.12) = 207 - 207.36 = -0.36 m/s. Since the velocity is negative, it means the cannonball is at its maximum height at this time. Therefore, the maximum vertical velocity of the cannonball is approximately 200 m/s.

The statement "Speed is the magnitude of velocity" is correct because speed refers to the rate at which an object covers distance, while velocity refers to the rate at which an object changes its position in a specific direction. Speed is a scalar quantity, meaning it only has magnitude, while velocity is a vector quantity, meaning it has both magnitude and direction. Therefore, speed is the magnitude of velocity, indicating how fast an object is moving regardless of its direction.

The average velocity of a particle is calculated by finding the change in position divided by the change in time. In this case, we are given the position function x(t) = t(t-5)^2, and we need to find the average velocity during the time interval [0,4]. To calculate this, we first find the change in position by plugging in the values of t=4 and t=0 into the position function and subtracting the results. Then, we divide the change in position by the change in time, which is 4-0=4. After performing the calculations, we find that the average velocity is 1.

Acceleration is the time derivative of the rate of change of position. This means that acceleration measures how quickly the rate of change of position is changing over time. It indicates how fast an object's velocity is changing.

The correct answer is Graph C is the position, Graph B is the velocity, Graph A is the acceleration. This is because the position graph (Graph C) shows the object's displacement over time, the velocity graph (Graph B) shows the rate at which the position is changing, and the acceleration graph (Graph A) shows the rate at which the velocity is changing.

The particle's acceleration at t=2 can be found by taking the second derivative of the position function with respect to time. Taking the derivative of x(t) = t(t-5)^2 twice, we get x''(t) = 2(t-5) - 2(t-5) = -8. Therefore, the particle's acceleration at t=2 is -8.

The velocity function is the integral of the acceleration function. By integrating a(t) = 3t^2 + t - 3, we can find the velocity function. The integral of 3t^2 is t^3, the integral of t is (1/2)t^2, and the integral of -3 is -3t. Combining these integrals, we get v(t) = t^3 + (1/2)t^2 - 3t + C, where C is a constant of integration. Since the question asks for which functions could represent the velocity function, any function of the form v(t) = t^3 + (1/2)t^2 - 3t + C would be a valid answer. Therefore, A(t) and F(t) could both represent the velocity function.

The maximum vertical speed of the cannonball can be determined by finding the derivative of the height function with respect to time and setting it equal to zero. Taking the derivative of h(t) = 10 + 207t - 4.9t^2 gives us v(t) = 207 - 9.8t. Setting v(t) equal to zero and solving for t, we find t = 21.12 seconds. Substituting this value back into the velocity function, we get v(21.12) = 207 - 9.8(21.12) = 207 - 207.376 = -0.376 m/s. Since the velocity is negative, it means the cannonball is at its highest point and about to start falling. Therefore, the maximum vertical speed of the cannonball is 207 m/s.

By taking the derivative of velocity with respect to time, we can find the instantaneous acceleration. This is because acceleration is defined as the rate of change of velocity with respect to time.

We can also find the maximum velocity by taking the derivative of velocity. The maximum velocity occurs at the point where the derivative of velocity is equal to zero or undefined.

Finally, by analyzing the derivative of velocity, we can determine intervals of increasing or decreasing velocity. If the derivative is positive, the velocity is increasing, and if the derivative is negative, the velocity is decreasing.

Therefore, the correct answers are instantaneous acceleration, maximum velocity, and intervals of increasing or decreasing velocity.