A record of travel along a straigt path is as follows:
a) Start from rest with constant acceleration of 3.06m/s2 for 19.8s;

b) Constant velocty for the next .845 min;

c) Constant negative acceleration of -9.12m/s2 for 4.46s.

What is the total displacement x for the compete trip?

The distance traveled for each of these segments can be calculated using the formulas:

v = v0 + a*t;
d = d0 + v0*t + 1/2 a*t^2

Where d0 is the initial displacement at time t=0, and v0 is the initial velocity.

Post back with the answer you get.

I got 1069.26m, which is wrong. I tried finding the velocity for 1 and 2 with the first equation, and then found the distance and added them together. But it didn't work.

I'll help you through the first two steps, and then you can take it from there.

The distance traveled in the first segment is:

D1 = 1/2 a t^2 = 1/2* 3.06m/s^2*(19.8m/s)^2 = 599.8 m

The velocity at the end of the first segment:
V1 = at = 3.06m/s^2 * 19.8 s = 60.6 m/s

For the second segment, the distance traveled is: v1*t2 = 60.6 m/s * .845 min * 60 sec/mim = 3071.8 m.

So the total displacement after the second segment is 599.8 m + 3071.8 m = 3671.6 m.

Now, can you figure out the distance traveled in the third segment and add it to this?