Motion diagram w/ questions

[rcmango]

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.

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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!

 

[Chi Meson]

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.

 

[kyle8921]

Hi there! I can see that you have put in a lot of effort to understand the motion diagram. Let me try to help clarify some of your questions.

b. The notation used is just a way to express the velocity as a vector, with the i^ and j^ representing the x and y components of the velocity. You can also express it as 4.5 m/s, but using the vector notation can be helpful in understanding the direction of the velocity.

c. To get the average speed, you need to divide the total distance traveled by the total time taken. In this case, the total distance traveled is 4.5 m + 6 m = 10.5 m, and the total time taken is 20 seconds. So, 10.5 m / 20 s = 0.525 m/s. This is the average speed for the 20 seconds.

d. You are correct, you can use the Pythagorean theorem to find the magnitude of the displacement vector. The notation used is just another way to express the displacement vector, with the i^ and j^ representing the x and y components of the vector. You can also use inverse tan to find the angle, but it might be easier to use the notation given to find the angle.

f. Velocity final (Vf) is the velocity at the end of the time period, and velocity initial (Vi) is the velocity at the beginning of the time period. In this case, Vf is the velocity at 20 seconds, and Vi is the velocity at 0 seconds. The notation is used to represent the change in velocity (Vf - Vi) divided by the time taken (20 seconds).

g. The acceleration is calculated using the change in velocity (Vf - Vi) divided by the time taken (20 seconds). In this case, the change in velocity is (3 m/s i^ + 1.5 m/s j^) - (1 m/s j^) = 2 m/s i^ + 1.5 m/s j^. Dividing this by 20 seconds gives an acceleration of 0.1 m/s^2 i^ + 0.075 m/s^2 j^. The angle is 0 degrees because the acceleration is in the direction of the x-axis, which is represented by the i^ vector.

I hope this helps clarify your questions. Keep up the good work in trying to understand