[lillib123]

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?

[ebaines]

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.

[lillib123]

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.

[ebaines]

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?