Given the graph of acceleration vs time (picture), draw a graph of position vs time and velocity vs time. The initial conditions for the motion are z(t=0) = 2.00m and vz(t=0) = -1.50 m/s. Both graphs should be numerically correct, and show how the values were obtained.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Given the graph of acceleration vs time (picture), draw a graph of position vs time and velocity vs time. The initial conditions for the motion are z(t=0) = 2.00m and vz(t=0) = -1.50 m/s. Both graphs should be numerically correct, and show how the values were obtained.

Acceleration vs. Time
6.
4.
3.
0.
-1
-2
-3
0.5
1
1.5
2.5
3.5
t (s)
az (m/s²)
2.
Transcribed Image Text:Acceleration vs. Time 6. 4. 3. 0. -1 -2 -3 0.5 1 1.5 2.5 3.5 t (s) az (m/s²) 2.
Expert Solution
Step 1

Given information:

The initial position of the object z = 0

The initial velocity of the object vz = -1.50 m/s

Now, consider given graph we can see that the acceleration is decreasing from 0-1. s and remains constant from 1-3 sec. Let us calculate the slope of the acceleration from 0-1 sec as shown:

dadt=-2-01-0=-2da=-2dtIntigarate on the both sides we get:da=-2dta=-2t+cfrom the graph we can see that at t=0; a=0, at t=1; a=-2; substituting the two values we get:a0-1=-2tand we can see that the acceleration is constant from 1-3 s, we can write this as:a1-3=-2 m/s2

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