How Do You Calculate Motion Diagram Parameters for Constant Velocity?

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The discussion focuses on calculating motion diagram parameters for an object moving at constant velocities. The velocities for the first and second intervals are determined as 1.0 m/s in the j direction and 3.0 m/s in the i direction plus 1.5 m/s in the j direction, respectively. Average speed is calculated as 2.43 m/s, which is not simply the average of the two speeds but the total distance divided by total time. Displacement is expressed using unit vectors, with a magnitude of 42.4 m and an angle of 45 degrees. The discussion emphasizes the importance of using vector notation for clarity in physics calculations.
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Having lots of trouble understanding this entirely. there is a simple diagram attached too, here it is.

information:
A motion diagram for Olivia is created by indicating her position every 10.0s on a grid of squares that are 15.0m on a side. Olivia jogs at a constant velocity for the first 10.0s shown, and then jogs at a different, constant velocity for the second 10.0s.

Questions:

(a) What is her velocity for the first 10.0s?
(b) What is her velocity for the second 10.0s?
(c) What is her average speed for the 20.0s?
(d) what is her displacement for 20.0s? give answer as unit vectors i^ j^ and as a magnitude and a direction.
(e) what is her average velocity for the 20.0s?
(f) how to determine the direction of the acceleration?
(g) what is her average acceleration for the 20.0s? give answer in unit vectors i^ and j^ and as a magnitude and a direction.

answers:
a == 1.0 m/s j^
b == 3.00 ms/s i^ + 1.50m/s j^
c == 2.43 m/s
d == D = 30.0 m i^ + 30m j^, D = 42.4 m, theta = 45.0 degress
e == v = 1.50m/s i^ + 150 m/s j^
f == a = points right, (Vf - Vi)/(t) .. so the direction of a is in the direction of Vf - Vi.
g == a = 0.150 m/s^2 i^, a = 0.150 m/s^2, theta = 0.0 degrees.

==

alright, I'm having trouble with most of these.
heres what I've done.
a. is fine.

b. i see the answer, but must i use this notation is there another way to express this answer? is 4.5 m/s wrong?

c. how did we get 2.43 m/s? if we take the distance/time from 0 to 1 i get 1.5 and then from 1 to 2, i get 3, which is 4.5, so maybe 4.5 / 2 is 2.25? help please.

d. again like b. I understand the notation, but are we only using this notation to explain the displacement vector from 0 to 2? i can use pathagorean theorem for D length, and inverse tan for the angle.

e. is fine. its just double the velocity for a because its 20 seconds.

f. confused, because we've been expressing the velocity answers sometimes as vector notation?? so what exactly is velocity final, and velocity initial.

g. not sure why this is .15 because the acceleration goes from 1 to one corner below 2.
so why use 1.50m/10s and not 3.0m/10s, also have no idea on the angle being 0.

please help. thankyou!
 

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b) 4.5 is dead wrong. the i^ j^ notation is another way to state the "x" and "y" componants. The other way of stating the answer is to give the net magnitude of the velocity, which would be through the pythagorean theorem, 3.35 m/s, thenthe direction is found using trigonometry. But you are asked to keep things in the form of x and y components (i^ is x).

c) average velocity is total distance divided by time. IT will not be the average of the two speeds.

d) Oh good, you know that. The "unit vector" method (i^, j^, K^, for x, y, z) is preferred in advanced physics. get used to it.

f & g) treat your x and y seperately, as though they were from completey different things. look at your final x velocity and compare it to you initial x velocity. Divide that difference by the total time, and you have your x ( i^ ) aceleration.
 
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