A record of travel along a straight path is as follows:

1. Start from rest with constant acceleration of 2.77 m/s^2 for 15 s
2. Constant velocity for the next 2.05 min
3. Constant negative acceleration -9.47 m/s^2 for 4.39 s.

(a) What was the total displacement for complete trip?
(b) What were the average speeds for legs 1,2,and 3 of the trip as well as for the complete trip.

Answers in the book: (a) 5510 m (b) 20.8 m/s, 41.6 m/s, 20.8 m/s, 38.7 m/s

Just do it one step at a time and add the distances traveled. During part 1, the distance travelled is

(1/2) a t^2 = (1/2)(2.77)(225) = 311.6 m
The final speed attained is (2.77)(15) = 41.55 m/s. The average speed duering that interval is half of that, or 20.8 m/s
During part 2, the distance traveled is
41.55 m/s*2.05 min*60 sec/min = 5110 m/s
During part 3, the speed drops from 41.55 m/s to
41.55 - (9.47)(4.39) = 0
The average speed during this interval is 41.55/2 - 20.78 m/s and the distance travelled is 20.78*4.39 = 91.2 m/s
Total distance travelled = 312 + 5110 + 91 = 5513 m. In your book's answer, they rounded off some of the numbers differently. You can only trust the first three significant figures, hence the 5510 m answer.

How would I get the average speed for the complete trip? (38.7 m/s)

To find the total displacement for the complete trip, we need to find the displacements for each leg and then add them up.

Leg 1: Start from rest with constant acceleration for 15 s.
To find the displacement during this leg, we can use the equation:

d = v0t + (1/2)at^2

Where d is the displacement, v0 is the initial velocity, t is the time, and a is the acceleration.

Given:
v0 = 0 (since we start from rest)
a = 2.77 m/s^2
t = 15 s

Plugging in the values, we get:

d1 = 0(15) + (1/2)(2.77)(15^2)
= 0 + (1/2)(2.77)(225)
= 0 + (1.385)(225)
= 307.875 m

So, the displacement for leg 1 is 307.875 m.

Leg 2: Constant velocity for 2.05 min.
Since there is no change in velocity during this leg, the displacement is simply the product of velocity and time.

Given:
v = constant velocity
t = 2.05 min = 2.05 * 60 = 123 s

To find the displacement, we multiply the velocity by the time:

d2 = v * t

Unfortunately, the velocity (v) is not given, so we cannot find the displacement for leg 2. However, we can still find the average speed for this leg, which is the total distance traveled divided by the time taken.

Leg 3: Constant negative acceleration for 4.39 s.
Using the same equation as in leg 1, we can find the displacement during this leg.

Given:
v0 = constant velocity
a = -9.47 m/s^2
t = 4.39 s

Plugging in the values, we get:

d3 = v0(4.39) + (1/2)(-9.47)(4.39^2)
= v0(4.39) + (-4.735)(4.39^2)
= 4.39v0 - 0.165(19.3721)
= 4.39v0 - 3.185
= 4.39v0 - 3.185

Again, the constant velocity (v0) is not given, so we cannot find the displacement for leg 3. However, we can find the average speed for this leg.

Now, to find the total displacement for the complete trip, we add up the displacements from all three legs:

Total displacement = d1 + d2 + d3

Since d2 and d3 cannot be directly calculated without the values of the velocities, we cannot find the exact total displacement. It seems there might be an error in the answers in the book for total displacement.

To find the average speeds for legs 1, 2, and 3 as well as the complete trip, we use the formula:

Average speed = Total distance / Total time

The total distance for each leg is the same as the displacement, since the path is linear.

Average speed for leg 1:
Given time for leg 1 is 15 s, so the average speed is:

Speed1 = d1 / t1
= 307.875 / 15
= 20.525 m/s

Average speed for leg 2:
Since we don't have a displacement value for leg 2, we cannot calculate the average speed for this leg.

Average speed for leg 3:
Given time for leg 3 is 4.39 s, so the average speed is:

Speed3 = d3 / t3
= (4.39v0 - 3.185) / 4.39

Again, since the velocity (v0) is not given, we cannot find the average speed for leg 3.

For the complete trip, the average speed can be calculated using the total displacement (which is not given accurately) and the total time taken for the complete trip.

Average speed for the complete trip:
Given time for the complete trip is 15 + 123 + 4.39 = 142.39 s (sum of the times for each leg), so the average speed is:

Speed_complete_trip = Total displacement / Total time
= (d1 + d2 + d3) / (t1 + t2 + t3)
= (not calculable without accurate values for d2 and d3)