Question

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s = 4t^2 + 3.

(a) Find the average velocity between t = 1 and t = 1 + h if:
(i) h = 0.1
(ii)ん = 0.01
(ili) h = 0.001
Enter the exact answers.
(i) When h = 0.1, the average velocity between t = 1 and t = 1 + his
i
m/sec.
(ii) When h = 0.01, the average velocity between + = 1 and t = 1 + his i
m/sec.
(iti) When h = 0.001, the average velocity between + = 1 and + = 1 + his i
m/sec.
(b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time t = 1.
Round your estimate to the nearest integer.
The instantaneous velocity appears to be
m/sec.

Answer 1

Explanation:

(a) To find the average velocity between t = 1 and t = 1 + h, we can use the formula:

average velocity = (change in distance) / (change in time)

We are given that the distance traveled by the particle is s = 4t^2 + 3, so the change in distance between t = 1 and t = 1 + h is:

s(1+h) - s(1) = [4(1+h)^2 + 3] - [4(1)^2 + 3] = 8h + 4h^2

The change in time is simply h. Therefore, the average velocity is:

(i) When h = 0.1:

average velocity = (8(0.1) + 4(0.1)^2) / 0.1 = 8.4 m/sec

(ii) When h = 0.01:

average velocity = (8(0.01) + 4(0.01)^2) / 0.01 = 8.04 m/sec

(iii) When h = 0.001:

average velocity = (8(0.001) + 4(0.001)^2) / 0.001 = 8.004 m/sec

(b) To estimate the instantaneous velocity of the particle at time t = 1, we can take the limit of the average velocity as h approaches 0. That is:

instantaneous velocity = lim(h -> 0) [(8h + 4h^2) / h]

Using L'Hopital's rule, we can take the derivative of the numerator and denominator with respect to h:

instantaneous velocity = lim(h -> 0) [8 + 8h / 1]

instantaneous velocity = 8 m/sec

Therefore, the estimated instantaneous velocity of the particle at time t = 1 is 8 m/sec (rounded to the nearest integer).