Question

An object's position is given by x=bt+ct3, where b=1.50m/s and c=0.640m/s3. to study the limiting process leading to the instantaneous velocity, calculate the object's average velocity over time intervals from 1.00 s to 3.00 s.

Answer 1

To calculate the average velocity of the object over a given time interval, we can use the equation for position x and find the values of x at the start and end of the interval. Then, we divide the displacement by the time interval to calculate the average velocity. In this case, the average velocity from 1.00 s to 3.00 s is 10.915 m/s.

Step-by-step explanation:

To calculate the object's average velocity over the time interval from 1.00 s to 3.00 s, we first need to find the values of x(1.00 s) and x(3.00 s) using the given equation for position x.

Plugging in t = 1.00 s into x=bt+ct^3, we have x = (1.50)(1.00) + (0.640)(1.00)^3, which gives us x(1.00 s) = 2.14 m.

Similarly, plugging in t = 3.00 s, we have x = (1.50)(3.00) + (0.640)(3.00)^3, which gives us x(3.00 s) = 23.97 m.

The average velocity is then calculated by taking the displacement and dividing it by the time interval: average velocity = (x(3.00 s) - x(1.00 s)) / (3.00 s - 1.00 s).

Substituting the values we found, we have average velocity = (23.97 m - 2.14 m) / (3.00 s - 1.00 s) = 10.915 m/s.

Answer 2

x=bt + ct3, b=1.50m/s, c=0.640m/s3.
Average velocity = distance / time
ti = 1.00, tf = 3.00
xi = (1.50)(1.00) + (0.640 )(1.00)^3 = 1.5 + 0.640 = 2.14 m
xf = = (1.50)(3.00) + (0.640)(3.00)^3 = 4.5 + 27(17.28) = 21.78 m
Ave Velocity = (21.78 - 2.14) / (3 - 1) = 9.82 m/s