Physics Speed,Distance,Velocity Homework *HELP*

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SUMMARY

The discussion centers on solving a physics homework problem involving a car's acceleration, speed, and distance over specified time intervals. Key calculations include determining the car's speed at t = 4 seconds, total distance traveled in the first 5 seconds, and average speed from t = 5 seconds to t = 9 seconds. The area under the acceleration-time (a-t) graph is crucial for calculating velocity changes, while the area under the velocity-time (v-t) graph provides displacement. The user Monty successfully calculated the car's speed at various time points but struggled with distance calculations, particularly when the slope of the graph changes direction.

PREREQUISITES
  • Understanding of kinematics, specifically acceleration, velocity, and displacement.
  • Ability to calculate areas under graphs, particularly rectangles and triangles.
  • Familiarity with constructing and interpreting velocity-time (v-t) graphs.
  • Knowledge of basic algebra for solving equations related to motion.
NEXT STEPS
  • Learn how to calculate areas under graphs for varying slopes in kinematics.
  • Study the principles of average speed and how to apply them in different scenarios.
  • Explore more complex kinematic equations and their applications in real-world problems.
  • Practice constructing and analyzing v-t graphs for various motion scenarios.
USEFUL FOR

Students studying physics, particularly those tackling kinematics problems involving acceleration, velocity, and displacement. This discussion is beneficial for anyone needing clarification on graph interpretation and area calculations in motion analysis.

BraindeadX64
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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
 
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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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