Question

A box proceeds along the x-axis and the figure below shows a record of its velocity as a function of time. Every grid line along the vertical axis corresponds to 2.00 m/s and each gridline along the horizontal axis corresponds to 0.500 s. (Enter your answer in m/s^2. Indicate the direction with signs of your answers. Note that t=0 at the intersection of the axes.) a) Determine the average acceleration of the box in the time interval t=0 to t=2.50 s. b) Determine the average acceleration of the box in the time interval t=2.50 s to t=7.50 s. c) Determine the average acceleration of the box in the time interval t=0 to t=10.0 s

Answer 1

The average acceleration of an object where the velocity is given as a function of time is computed with the change in velocity over change in time formula. For the velocity function v(t) = 2 m/s + 0.25m/t, the acceleration is the derivative of velocity, leading to a(t) = -0.25m/t². The acceleration at t = 2.0 s is -0.0625 m/s² and at t = 5.0 s is -0.0100 m/s².

Step-by-step explanation:

To calculate the average acceleration of an object when given its velocity as a function of time, we use the following formula:

a = (vf - vi) / (tf - ti)

Where:

• vf is the final velocity
• vi is the initial velocity
• tf is the final time
• ti is the initial time

For the velocity function v(t) = A + Bt¯¹, where A = 2 m/s and B = 0.25 m, the acceleration is the derivative of velocity with respect to time. Therefore, the acceleration a(t) = -Bt¯² at any given time. At t = 2.0 s and t = 5.0 s:

a(2.0 s) = -0.25 m / (2.0 s)² = -0.0625 m/s²

a(5.0 s) = -0.25 m / (5.0 s)² = -0.0100 m/s²

To find the position, one would integrate the velocity function from the initial time to the time of interest. However, specifics are required for precise integration.