[1D kinematics] Car problem: Solving problem with a graph

In summary, a car starts from rest and goes through three stages of motion: uniform acceleration for 200m, constant speed for 160m, and deceleration to rest in 50m. The total time of the trip is 33s. Using the graphing method and kinematics equations, the constant speed portion can be determined by finding the area of the trapezoid on the v versus t graph.
  • #1
hiuting
23
0

Homework Statement


A car starts from rest and accelerates uniformly for 200m. It moves at constant speed for 160m and then decelerates to rest in 50m. The whole trip takes 33s. How long did it move at constant speed? (hint: draw the v versus t graph).


Homework Equations


kinematics equations
and the graph is used


The Attempt at a Solution


I drew the graph which looks like a trapezoid... but I don't know what to do after that.
 
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  • #2
Delightful to see your teacher using the graphing method instead of just formulas all the time! Do you know how to put the distances on that graph? Got the constant speed level marked with a "v" or something? The method usually involves writing formulas for the areas. Once you have them written up you'll probably see how to finish it.
 
  • #3


Great job on drawing the v versus t graph! This is a very useful tool in solving kinematics problems. To solve this problem, we can use the kinematics equations and the information given in the graph. Since the car starts from rest, we know that its initial velocity is 0 m/s. Using the area under the graph, we can find the total distance traveled by the car, which is 410m. We also know that the total time taken is 33s.

To find the constant speed portion of the trip, we can use the equation v = d/t, where v is the constant speed, d is the distance traveled, and t is the time taken. Using the values we have, we get v = 160m/13s = 12.31 m/s. This means that the car was moving at a constant speed of 12.31 m/s for 13 seconds.

We can also use the area under the graph to find the acceleration and deceleration of the car. The area of the trapezoid represents the change in velocity (from 0 m/s to 12.31 m/s and back to 0 m/s) over a certain time interval. Using the formula for the area of a trapezoid, we can find the acceleration and deceleration to be 0.75 m/s^2.

In summary, the car moved at a constant speed of 12.31 m/s for 13 seconds. This can also be confirmed by calculating the total distance traveled during this portion of the trip (160m) and dividing it by the constant speed (160m/12.31 m/s = 13s).
 

1. How do I solve a car problem using a graph?

To solve a car problem using a graph, you will need to plot the car's position over time on a graph. The x-axis will represent time and the y-axis will represent the car's position. Then, you can use the slope of the line on the graph to determine the car's velocity.

2. What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement divided by the total time, while instantaneous velocity is the velocity at a specific moment in time. In other words, average velocity gives an overall picture of the car's movement, while instantaneous velocity shows the car's speed at a specific point in time.

3. How do I calculate acceleration from a position vs. time graph?

To calculate acceleration from a position vs. time graph, you will need to find the slope of the velocity vs. time graph. Acceleration is the change in velocity divided by the change in time, so the slope of the velocity vs. time graph will give you the acceleration.

4. Can I use a position vs. time graph to determine the car's displacement?

Yes, you can use a position vs. time graph to determine the car's displacement. The displacement is equal to the area under the graph, which can be found by dividing the graph into smaller rectangles and adding their areas together.

5. What is the difference between distance and displacement?

Distance is the total length of the path traveled, while displacement is the straight-line distance between the starting and ending points. In other words, distance is the actual distance traveled, while displacement is the change in position.

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