Question

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, where distance is measured in meters and time in seconds. Find the approximate value of the average velocity (accurate up to three or more decimal places) in the given time intervals. TL between t1-5 s and t2-8 s: between t1-4 s and t2-9 s: between t1-1 s and 12-7s: 8 TTL 8 TTL 8

Answer 1

The average velocity between two time points is calculated by taking the difference in the object's vertical position at these times, divided by the time elapsed. This is done by evaluating the position equation L(t) at the specified times and using these values to find the average velocity.

Step-by-step explanation:

To find the average velocity of an object moving along a vertical line with its vertical position given by the equation L(t) = 2t^3 + t^2 - 5t + 1, we use the formula for average velocity, which is the total displacement divided by the total time, or v_{avg} = (L(t_2) - L(t_1)) / (t_2 - t_1). For the time intervals given, we would separately evaluate L(t) at the two specified times for each interval and then compute the average velocity.

As an example, to calculate the average velocity between t_1 = 5 s and t_2 = 8 s, we find the positions L(5) and L(8) and then compute (L(8) - L(5)) / (8 - 5).

For accurate results, the values of L(t) should be computed using a calculator or software tool that can handle polynomial expressions to at least three decimal places of accuracy.

Answer 2

The average velocity is

`266(m)/(s),274(m)/(s)` and
`117(m)/(s)` respectively.

Step-by-step explanation:

Let's start writing the vertical position equation :

`L(t)=2t^(3)+t^(2)-5t+1`

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

`V_(avg)=(Displacement)/(Time)` = Δx / Δt =
`(x2-x1)/(t2-t1)`

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

`t2-t1=8s-5s=3s`

For the position variation we use the vertical position equation :

`x2=L(8s)=2.(8)^(3)+8^(2)-5.8+1=1049m`

`x1=L(5s)=2.(5)^(3)+5^(2)-5.5+1=251m`

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

`(798m)/(3s)=266(m)/(s)`

For the second time interval :

t1 = 4 s → t2 = 9 s

`x2=L(9s)=2.(9)^(3)+9^(2)-5.9+1=1495m`

`x1=L(4s)=2.(4)^(3)+4^(2)-5.4+1=125m`

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

`(1370m)/(5s)=274(m)/(s)`

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

`x2=L(7s)=2.(7)^(3)+7^(2)-5.7+1=701m`

`x1=L(1s)=2.(1)^(3)+1^(2)-5.1+1=-1m`

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

`(702m)/(6s)=117(m)/(s)`