Question

Consider the velocity data for the car shown in table 5.1. Find lower and upper estimates for the distance it travels between t=2 and t=8.

lower =
upper =
​
Write the difference between these as the following product: difference =(Δt)(Δv)=1, Note: You can earn partial credit on this problem.

Answer 1

To find the lower and upper estimates for the distance traveled between t=2 and t=8, we can use the formula: d = vavg * t. Lower estimate = 20m, upper estimate = 160m, difference between estimates = 80m.

Step-by-step explanation:

To find the lower and upper estimates for the distance traveled between t=2 and t=8, we can use the formula: d = vavg * t, where d is the distance, vavg is the average velocity, and t is the time interval.

To find the lower estimate, we use the initial velocity (v1) and the time at t=2. To find the upper estimate, we use the final velocity (v2) and the time at t=8. The difference between these estimates can be calculated as Δd = (v2 - v1) * (t2 - t1).

Let's look at the concrete example and calculate the lower and upper estimates:

Given data:

Time (s)Velocity (m/s)210820

Lower estimate: d = v1 * t1 = 10 m/s * 2 s = 20 m

Upper estimate: d = v2 * t2 = 20 m/s * 8 s = 160 m

Difference between estimates: Δd = (20 m/s - 10 m/s) * (8 s - 2 s) = 80 m

So, the lower estimate for the distance is 20 m, the upper estimate is 160 m, and the difference between the estimates is 80 m.

Answer 2

The lower and upper estimates for the distance it travels between t=2 and t=8 are 152 feet and 192 feet respectively. The difference between these as expressed as
`(\Delta t)(\Delta v) = (4)(10) = 40`.

To estimate the distance traveled by the car between t = 4 and t = 8 using lower and upper estimates, we can use the trapezoidal rule.

The formula for estimating the distance traveled using the trapezoidal rule with velocity data is:

`\[\text{Distance} \approx \text{Average Velocity} * \text{Time Interval}\]`

For the time interval t = 4 to t = 8, the average velocity can be estimated by taking the average of the velocities at t = 4 and t = 8.

Given:

- Velocity at t = 4 is 38 ft/sec.

- Velocity at t = 8 is 48 ft/sec.

- Time interval
`\(\Delta t = 8 - 4 = 4\)` seconds.

1. Lower Estimate:

Using the lower estimate, assume the velocity remains constant at
`\(t = 4\)` (38 ft/sec) throughout the interval from
`\(t = 4\) to \(t = 8\)`.

`\[\text{Distance (lower estimate)} = \text{Velocity at } t = 4 * \text{Time Interval}\ \\\ \[\text{Distance (lower estimate)} = 38 * 4 = 152 \text{ feet}\]`

2. Upper Estimate:

Using the upper estimate, assume the velocity remains constant at
`\(t = 8\)` (48 ft/sec) throughout the interval from
`\(t = 4\) to \(t = 8\)`.

`\[\text{Distance (upper estimate)} = \text{Velocity at } t = 8 * \text{Time Interval}\]`

`\[\text{Distance (upper estimate)} = 48 * 4 = 192 \text{ feet}\]`

Now, find the difference between the upper and lower estimates as a product of
`\(\Delta t\)` (time interval) and
`\(\Delta v\)` (difference in velocities).

`\(\text{Difference} = (\Delta t)(\Delta v) = (4)(48 - 38) = (4)(10) = 40\)`

Therefore, the difference between the upper and lower estimates for the distance traveled between t = 4 and t = 8 is 40 feet, which can be expressed as
`(\Delta t)(\Delta v) = (4)(10)`.

Table 5.1 is given below: