Anyone good at Physics???

Question

I've been working on Physics homework for the past 4 hours...I have finished all of the 35 problems except for this one:

14) A record of travel along a straight path is as follows:
(a) Start from rest with constant acceleration of 2.33 m/s for 14.4s;
(b) Constant velocity for the next 55.2 seconds;
(c) Constant negative acceleration of -9.76 m/s for 5.59s.
What was the total displacement X for the complete trip? answer in meters

Answer 1

The total displacement is going to be 2128.53 meters

I.) Xf=Xi +Vi(t)+1/2a*change in time squared
Xf=0+0+1/2(2.33)(14.4 squared)
Xf=241.75 meters
II.) Vf^2= Vi^2 + 2a(Xf-Xi)
Vf^2= 0+2(2.33)(241.57)
Vf=33.55
III.) 33.55 m/s * 55.2s=1851.96 meters
IV.) Use the first equation given:
Xf=0+33.55(5.59)+1/2(-9.76)(5.59)^2
Xf=35 meters
V.) add them together:
Xt=241.57+1851.96+35=2128.53

Answer 2

Are you using the following equation?

x = 1.2 * a * t^2 + v * t

x = displacement, meters
a = acceleration, meters per second squared
t = time, sec
v = velocity, meters per second

Do parts a, b, and c separately and add them together. Remember to make you units consistent and don't forget the sign on the acceleration in part c.

Answer 3

SORRY i can't help. i admire you for giving it a good try before asking for some help. good luck i hope some one reads your question who can help,

Answer 4

beats me!...sorry.