i dont understand need help ASAP!!!?

Question

I have a graph of velocity versus time for constant acceleration after looking at the graph I have to fill in a chart with four headings in each column: TIME (s), VELOCITY (m/s), ACCELERATION (m/s^2), AND DISPLACEMENT (m). There are 4 rows under time and velocity. In the column that says time here is whats given 0.0, 2.0, 5.0, and 10.0. I got the velocity by looking at the graph. Heres what I got. At time 0.0 the velocity is 1.0 at time 2.0 the velocity is 2.0 at time 5.0 the velocity is 3.5 and at time 10.0 I got 6.0 under the heading acceleration and displacement there are only 3 rows. Heres what I got:
At time 0.0, velocity 1.0 acceleration is 0.5, at time 2.0, velocity 2.0, the acceleration is 0.5 at time 5.0, velocity 3.5, the acceleration is 0.5. what I wanna know is does this sound correct? Do the velocity and acceleration make sense according to the time given? Also how would I calculate the displacement for these 3 is there a formula I use to calculate the displacement? Please help im having dificulty understanding this problem and I still have one more with a graph of velocity versus time for motion with variable acceleration. Please show ur work and explain thoroughly. Maybe drawing a table with the info I have given may help u understand the question if it doesn’t make sense in the way I wrote it.

Answer 1

Acceleration is change in velocity per unit time.

That is acceleration = change in velocity / (change in time)

In the chart, the third column is for acceleration. You are to fill up the third column using the graph.

In the first row, leave the third (acceleration) and fourth (distance) blank.

In the second row, we can fill up the acceleration column.

The change in time from zero to two is 2. (difference between the second and first row).

The change in velocity from one to two is 1.

Acceleration is ½ = 0.5.

For the third row,

The change in time is 5 - 2 = 3.

The change in velocity is 3.5 - 2 = 1.5

Acceleration is 1.5 / 3 = 0.5.

Similarly you can fill the other rows.

Do the same things, even if the acceleration varies.


The fourth column is filled up either using the graph or calculations.
Let the x- axis be the time axis and y- axis be the velocity axis.

To fill up the distance for the second row, find the area formed by the two vertical lines at t= 0 and t = 2 and the lines v = 1 and v= 2.

Or use the formula, distance = average velocity x time interval.
Average velocity of v=1 and v= 2 is (1 +2) / 2 = 1.5.
Time interval is (2 - 0) = 2.
Distance traveled is 1.5 x 2 = 3m.

For the third row, {(3.5 + 1) /2} x (5- 0) = 11.25 m.
Average velocity of v=1 and v= 3.5 is (1 +3.5) / 2 = 2 .25.
Time interval is (5-0) = 5
Distance traveled is 2.25 x 5 =11.25 m

Note that the distances are calculated from the start to the final time.

Answer 2

OK, you did it correctly, because at the beginning you had the phrase that the graph is for constant acceleration.
The displacement for the graph means the area beneath the graph curve. So you have to calculate the displacement (for constant acceleration) as the velocity at zero time plus velocity at some other time and after summing to multiply by the time interval and divided by two.
So, Dfi = (Vf + Vi)(Tf - Ti)/2;
and for the constant acceleration movement it means
D(T) = Vo + aT^2/2, as Vf = Vi + a(Tf - Ti)
For given problem
D(0) = 0, there is no need in the column,
D(2) = 3m
D(5) = 11.25 m
D(10) = 35 m (these are values obtained from the first formula,
D = (V(0) + V(T))*T/2, but we can obtain the same
results using the second formula,
D = V(0)T + aT^2/2, you can check)
When you have small period of time, the body, which moves with acceleration, overcomes distance as the average velocity multiplied by this period duration, and the average velocity is velocity at the beginning of the period plus the velocity at the end of the period divided by two. Taking into account that the velocity at the end of the period is the velocity at the beginning plus acceleration multiplied by the duration of the period, we obtain the algorithm for calculating the displacement.
Dfi = (Vi+Vf)(Tf-Ti)/2 = (Vi + Vi + a(Tf - Ti))(Tf - Ti)/2 =
= (2Vi + a(Tf - Ti))(Tf - Ti)/2 = Vi(Tf - Ti) + a(Tf - Ti)^2/2
This is the usual formula for accelerated motion.
In case the acceleration is not constant, you will have to divide the intervals so that on each interval you can calculate the parameters of velocity and displacement.

Answer 3

Your velocities do not look right. A constant acceleration results in a linear velocity with time, which is not what you show.

Answer 4

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Answer 5

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