physics acceleration and displacement problem?

Question

A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.05 m/s2 for 20.0 s
2. Constant velocity for the next 2.70 min
3. Constant negative acceleration 8.77 m/s2 for 4.73 s
(a) What was the total displacement for the complete trip?
m
(b) What were the average speeds for legs 1, 2, and 3 of the trip as well as for the complete trip?
m/s
m/s
m/s
m/s

Answer 1

Lets clarify:
- Total displacement = Position(final) - Position(initial)
- Total distance traveled = Sum of absolute values of incremental distances
Note: Toward the end of the trip the moving object slows to a stop and starts moving back the other way.

a)
Leg One:
v = at = (2.05)(20) = 41m/s
s = (1/2)a t^2 = (1/2)(2.05)(20)^2 = 410m
Leg Two:
s = vt = 41(2.70)(60) = 6642m
Leg Three:
(41m/s)/(8.77m/s^2) = 4.675s (slows to a stop in 4.675 sec.)
s = (1/2)at^2 = (1/2)(8.77)(4.675)^2 = 95.837m
4.73 - 4.675 = 0.05497 seconds remaining
(8.77)(0.05497) = 0.482m/s (speeds up to 0.482m/s the other way)
s = (1/2)at^2 = (1/2)(8.77)(0.05497)^2 = 0.01325m
Total displacement: (maintain signs of incremental displacements)
= 410 + 6642 + 95.837 - 0.01325 = +7147.82m

b) Average speed:
Leg one:
v = at = 2.05(20) = 41m/s
Leg Two:
41m/s
Leg Three:
total distance = 95.837 + 0.01325 = 95.85m
speed = total distance / time = (95.85)(4.675) = 20.503m/s
Complete trip:
Total distance traveled = 410 + 6642 + 95.837 + 0.01325 = 7147.85m
(Notice that total distance traveled is greater than total displacement.)
Total time = 20 + 162 + 4.73 = 186.73s
Average speed = 7147.85/186.73 = 38.279m/s

Answer 2

Acceleration could nicely be evaluated straight away. In an era of five.0 seconds there could desire to be many diverse values of acceleration. question could desire to the two state that the braking acceleration became consistent or that the effortless vector acceleration is to be stumbled on. Avg fee of acceleration = (eighty 5 - 5)/5 = sixteen m/s² DECELERATION ANS 1a course = S ANS 1b Displacement length = 5Vavg = 5(eighty 5+5)/2 = 5(40 5) = 225 m ANS 2a Displacement course = N ANS 2b

Answer 3

How to work out without Gravity ? Sorry though Engineer I forgot the formula. Now my formula is how to accelerate FREEDOM from life cycle.