The position of a particle along a straight-line path is defined by s=(t3−6t2−15t+7) ft, wheret is in seconds.A. Determine the total distance traveled when t = 8.3 s .B. What are the particle's average velocity at the time given in part A?C. What are the particle's average speed at the time given in part A?D. What are the particle's instantaneous velocity at the time given in part A?E. What are the particle's acceleration at the time given in part A?

First, let´s find at which time the particle changes direction. The sign of the instantaneous velocity indicates the direction of the particle. We will consider the right direction as positive. The origin of the frame of reference is located at s = 0 and t = 0 so that the particle at t = 0 is located 7 ft to the right of the origin.

The instaneous velocity (v(t)) of the particle is the first derivative of s(t):

v(t) = ds/dt = 3t² - 12t - 15

The sign of v(t) indicates the direction of the particle. Notice that at t = 0,

v(0) = -15. So, initially, the particle is moving to the left.

So let´s find at which time v(t) is greater than zero:

v(t)>0

3t² - 12t - 15>0

Solving the quadratic equation with the quadratic formula:

For every t > 5 s, v(t) > 0 (the other solution of the quadratic equation is -1. It is discarded because the time can´t be negative).

Then, the particle moves to the left until t = 5 s and, thereafter, it moves to the right.

To find the traveled distance at t= 8.3 s, we have to find how much distance the particle traveled to the left and how much distance it traveled to the right.

So, let´s find the position of the particle at t = 0, at t = 5 and at t = 8.3 s

s(t) = t³ - 6t² - 15t + 7

s(0) = 7 ft

s(5) = 5³ - 6 · 5² - 15 · 5 + 7 = -93 ft

s(8.3) = 8.3³ - 6 · 8.3² - 15 · 8.3 + 7 = 40.9 ft

So from t = 0 to t = 5, the particle traveled (93 + 7) 100 ft to the left, then from t = 5 to t = 8.3 the particle traveled (93 + 40.9) 134 ft to the right. Then, the total distance traveled when t = 8.3 s is (134 ft + 100 ft) 234 ft.

B) The average velocity (AV) is calculated as the displacement over time:

AV = Δs / Δt

Where:

Δs = displacement (final position - initial position).

Δt = elapsed time.

In this case:

final position = s(8.3) = 40.9 ft

initial position = s(0) = 7 ft

Δt = 8.3 s

So:

AV = (s(8.3) - s(0)) / 8.3 s

AV = (40.9 ft - 7 ft) / 8.3 s

AV = 4.1 ft/s

The average velocity of the particle is 4.1 ft/s (since it is positive, it is directed to the right).

C) The average speed is calculated as the traveled distance over time. The traveled distance at t = 8.3 s was already obtained in part A: 234 ft. Then, the average speed (as) will be:

as = distance / time

as = 234 ft / 8.3 s

as = 28 ft/s

The average speed at t = 8.3 s is 28 ft/s

D) The instantaneous velocity at any time t was obtained in part A:

v(t) = 3t² - 12t - 15

at t = 8.3 s

v(8.3) = 3(8.3)² - 12(8.3) - 15

v(8.3) = 92 ft/s

The instantaneous velocity at t = 8.3 s is 92 ft/s.

E) The particle acceleration at any time t, is obtained by derivating the velocity function:

v(t) = 3t² - 12t - 15

dv/dt = a(t) = 6t - 12

Then at t = 8.3 s

a(8.3) = 6(8.3) - 12

a(8.3) = 37.8 ft/s²

The acceleration of the particle at t = 8.3 s is 37.8 ft/s²