Physics Speed,Distance,Velocity Homework *HELP*

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Homework Help Overview

The discussion revolves around a physics problem involving a car's motion, specifically focusing on speed, distance, and velocity. The original poster presents a scenario where a car accelerates from rest, with a histogram depicting its acceleration over time. The problem requires calculations of speed and distance at various time intervals based on the acceleration data.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the area under the acceleration-time graph to determine changes in velocity and subsequently calculate distance. Some suggest constructing a velocity-time graph to aid in these calculations. There are attempts to apply formulas for velocity based on acceleration and time, but confusion arises regarding the correct interpretation of the graph and the calculations of distance.

Discussion Status

Several participants are actively engaging with the problem, sharing their calculations and expressing confusion about specific steps, particularly regarding how to handle negative slopes in the velocity-time graph. Some guidance has been offered about using the area under the graphs to find distances, but there is still uncertainty and a lack of consensus on how to proceed with the calculations.

Contextual Notes

Participants note the urgency of the homework deadline, indicating a time constraint that may be affecting their ability to clarify their understanding. There is also mention of potential confusion regarding the implications of negative slopes in the context of distance calculations.

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Homework Statement



A car is initially at rest on a straight road. The histogram below shows the car's acceleration along that road as a function of time.

plot.png


1. Calculate the speed of the car at t = 4 s.

2. Calculate the distance traveled during the first 5 s.

3. Calculate the distance traveled from t=10 s to t=14 s.

4. Calculate the car's average speed from t = 5 s to t=9 s.

Homework Equations



The area under the a-t graph is the change in the object's velocity. This problem involves very simple areas to calculate (rectangles and squares). You can then use this information to construct a v-t graph; the area under the v-t plot is the displacement (these areas will be rectangles and triangles).

Another option is to find the velocity at each time the acceleration changes; then use the velocity at the start of the interval and the velocity at the end of the interval to calculate the average velocity over each interval in which the acceleration is constant. The displacement over this interval is just the average velocity*time interval. You will need to find the displacement separately for each value of constant acceleration within your given time frame, the total displacement is just the sum of the individual displacements.

The average speed over a time interval can be calculated if you know the total distance traveled over that interval. Average speed=total distance/time interval.

V(1)= V(0) + acceleration*time
V(2)= V(1) + acceleration*time

The Attempt at a Solution



I don't even know where to start. This online class is killing me. Any input will be helpful.

Thank you,
Monty
 
Last edited:
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I think I am getting somewhere with

V(1)= V(0) + acceleration*time
V(2)= V(1) + acceleration*time

Not sure if this is correct
 
BraindeadX64 said:
The area under the a-t graph is the change in the object's velocity. This problem involves very simple areas to calculate (rectangles and squares). You can then use this information to construct a v-t graph; the area under the v-t plot is the displacement (these areas will be rectangles and triangles).

Make a v-t graph for question number one. V-t graph is extremely helpful and is necessary for the completion of your questions.
 
I got the first answer

V(1)= V(0) + acceleration*time v1=2

v(2) = v(1) + (a*t) v2 = 8

v(3) = v(2) + (a*t) v3 = 17

v(4) = v(3) + (a*t) v4 = 25

it was 25m/s^2 but now I am confused with the 2nd one, How do I know the distance traveled?
 
Make a v-t graph and use the area to find your distances.

On a v-t graph the area is basically v x t, so you will get d.
 
JustinLiang said:
Make a v-t graph and use the area to find your distances.

On a v-t graph the area is basically v x t, so you will get d.

I did that and got 72 meters but it was incorrect.
 
Hmmm I don't remember this too well but if I recall correctly, your slope can go up or down, therefore you won't be adding all the distances. At times you may need to subtract distances as you are moving backwards.
 
JustinLiang said:
Hmmm I don't remember this too well but if I recall correctly, your slope can go up or down, therefore you won't be adding all the distances. At times you may need to subtract distances as you are moving backwards.

thats where I am confused it sloped down 5 and I don't know how to calculate that.
 
So when you the slope goes down you calculate the area under it and subtract it to the distance you have traveled thus far.
 
  • #10
JustinLiang said:
So when you the slope goes down you calculate the area under it and subtract it to the distance you have traveled thus far.

that doesn't make sense to me
 
  • #11
any help due in 1hr 20 mins

I got the velocitys of each pt

V(1)= V(0) + acceleration*time v1=2

v(2) = v(1) + (a*t) v2 = 8

v(3) = v(2) + (a*t) v3 = 17

v(4) = v(3) + (a*t) v4 = 25

v(5) = v(4) + (a*t) v5 = 20

v(6) = v(5) + (a*t) v6 = 32

v(7) = v(6) + (a*t) v7 = 46

v(8) = v(7) + (a*t) v8 = 54

v(9) = v(8) + (a*t) v9 = 36

v(10) = v(9) + (a*t) v10 = 26

v(11) = v(10) + (a*t) v11 = 26

v(12) = v(11) + (a*t) v12 = 26

v(13) = v(12) + (a*t) v13 = 26

v(14) = v(13) + (a*t) v14 = 26
 
  • #12
Heres the VT I made

Ez.jpg
 

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