*VELOCITY OF A CAR* Suppose the distance s (in feet) covered by a car moving along a straight road after t sec is given by the function s = f(t) = 2t^2 + 48t.

a. Calculate the average velocity of the car over the time intervals [20,21], [20, 20, 20.1], and [20, 20.01].

b. Calculate the (instantaneous) velocity of the car when t = 20.

c. Compare the results of part (a) with that of part (b).

*Can someone please help me? I have been trying to do my homework for some time now and I am stuck on this problem. I tried to work it out, but I did not understand what to substitute in for “t”. I used the average velocity formula and did f(t+h)-f(t) divided by h …. And I got: 4t + 48. Somehow, I think this is way wrong! Can you tell me where I went wrong here and how to do this problem!?? Pleaseee!

any help would be great!!! :)

The instantaneous velocity is indeed 4t+48

However, the average velocity is the average of the instaneous velocties at two times. Example, in a, times 20,and times 20.1
v(20)=4t+48=128
v(20.1)=128.4

average: 128.2

do you have the dike too?

Sure! I'd be happy to help you with this problem step-by-step.

a. To calculate the average velocity of the car over the time intervals [20,21], [20, 20.1], and [20, 20.01], we need to use the average velocity formula. The average velocity is given by the change in distance divided by the change in time.

For the interval [20,21]:
Average velocity = (f(21) - f(20)) / (21 - 20)
= (2(21)^2 + 48(21) - (2(20)^2 + 48(20))) / (1)
= (882 + 1008 - 848) / 1
= 1042 feet/second

For the interval [20, 20.1]:
Average velocity = (f(20.1) - f(20)) / (20.1 - 20)
= (2(20.1)^2 + 48(20.1) - (2(20)^2 + 48(20))) / (0.1)
= (808.2 + 964.8 - 848) / 0.1
= 924 feet/second

For the interval [20, 20.01]:
Average velocity = (f(20.01) - f(20)) / (20.01 - 20)
= (2(20.01)^2 + 48(20.01) - (2(20)^2 + 48(20))) / (0.01)
= (808.0402 + 964.08 - 848) / 0.01
= 916.08 feet/second

b. To calculate the instantaneous velocity of the car when t = 20, we can find the derivative of the function f(t) with respect to t. The derivative will give us the instantaneous velocity at any given point.

f(t) = 2t^2 + 48t

f'(t) = 4t + 48

When t = 20,
Instantaneous velocity = f'(20) = 4(20) + 48 = 128 feet/second

c. Comparing the results of part (a) with part (b), we see that the average velocities calculated for the different time intervals [20,21], [20, 20.1], and [20, 20.01] are different from the instantaneous velocity at t = 20.

The average velocities give us the average rate of change of distance over the given time intervals, while the instantaneous velocity gives us the rate of change at a specific point in time. Since velocity can change over time, the average velocity over an interval may not be the same as the instantaneous velocity at a single point.

Hope this helps! Let me know if you have any further questions.

Sure, I can help you with this problem. Let's go step by step to find the answers.

a) To calculate the average velocity over the time interval [20,21], [20, 20, 20.1], and [20, 20.01], we need to use the average velocity formula:

Average velocity = (change in distance) / (change in time).

For the interval [20, 21]:
The distance covered at time t1 = 20 is given by f(20) = 2(20)^2 + 48(20) = 800 feet.
The distance covered at time t2 = 21 is given by f(21) = 2(21)^2 + 48(21) = 882 feet.

The change in distance is 882 - 800 = 82 feet.
The change in time is 21 - 20 = 1 second.

Therefore, the average velocity over the interval [20, 21] is 82 feet/1 second = 82 feet/second.

For the interval [20, 20.1]:
The distance covered at time t1 = 20 is still f(20) = 800 feet.
The distance covered at time t2 = 20.1 is given by f(20.1) = 2(20.1)^2 + 48(20.1) ≈ 802.008 feet.

The change in distance is 802.008 - 800 = 2.008 feet.
The change in time is 20.1 - 20 = 0.1 second.

Therefore, the average velocity over the interval [20, 20.1] is approximately 2.008 feet/0.1 second ≈ 20.08 feet/second.

For the interval [20, 20.01]:
The distance covered at time t1 = 20 is still f(20) = 800 feet.
The distance covered at time t2 = 20.01 is given by f(20.01) = 2(20.01)^2 + 48(20.01) ≈ 800.8002 feet.

The change in distance is 800.8002 - 800 = 0.8002 feet.
The change in time is 20.01 - 20 = 0.01 second.

Therefore, the average velocity over the interval [20, 20.01] is approximately 0.8002 feet/0.01 second ≈ 80.02 feet/second.

b) To calculate the instantaneous velocity at t = 20, we need to find the derivative of the function f(t) with respect to t.

f'(t) = d/dt (2t^2 + 48t) = 4t + 48.

Substitute t = 20 into the derivative:

f'(20) = 4(20) + 48 = 80 + 48 = 128 feet/second.

Therefore, the instantaneous velocity at t = 20 is 128 feet/second.

c) Comparing the results of part (a) and part (b), you can see that the average velocity over the interval [20, 21] is 82 feet/second, which is different from the instantaneous velocity at t = 20, which is 128 feet/second.

It is important to note that average velocity is the change in distance divided by the change in time over an interval, while instantaneous velocity is the velocity at a specific moment in time. The average velocity considers the overall motion over an interval, whereas the instantaneous velocity focuses on the velocity at a specific point.

I hope this explanation helps you understand the problem better. Let me know if you have any further questions!