1.3 Iridov - Find Time Interval ( Kinematics )

In summary: I used the wrong formula for the area of a trapezoid. The correct formula is (a+b)/2 * h. In summary, the conversation discusses a problem with calculating the time of motion for a car that moves with varying acceleration and deceleration. The average velocity of the car is given, and the total time of motion is 25 seconds. The conversation also mentions using the area of a trapezoid to solve the problem and finding a mistake in the calculations.
  • #1
razored
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Thank you for taking the time to read this lengthy post.

A car starts moving rectilinearly, first with acceleration a=5m/s2(the initial velocity is 0) then uniformly, and finally, decelerating at the same rate a, comes to stop.The total time of motion equals 25 seconds. The average velocity during whole time is 72 km/h. How long does the car move uniformly?WARNING : 2 MB IMAGE

http://img26.imageshack.us/img26/1403/defaultx.png" All my equations appear to be correct; however, when I calculate the t_{o}, I get an imaginary answer. What did I miss or do wrong?
 
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  • #2
razored said:
Thank you for taking the time to read this lengthy post.

A car starts moving rectilinearly, first with acceleration a=5m/s2(the initial velocity is 0) then uniformly, and finally, decelerating at the same rate a, comes to stop.The total time of motion equals 25 seconds. The average velocity during whole time is 72 km/h. How long does the car move uniformly?

Maybe consider the area of a trapezoid?

That is what your V(t) graph represents.

You know that the integral of V(t) yields X(t)

They tell you what the area is ... avg V * total T = 20 * 25 = 500.

The base is 25 sec and total X is 500
So ...

(25+X)/2*H = 500

You also know that X = 25 - 2d
where d is the 2 periods of acceleration and deceleration.

And H = 5*d
where V = 5*t so ...

H = 5*(25 -X)/2

Combining things then yields:

5*(25 -X)/2*(25+X)/2 = 500

5/4*(252 - X2) = 500

I get a real result.
 
  • #3
Actually, I ignored the graph I drew when I solved the problem because I realized(on another piece of paper) that the area I was calculating (two triangles and rectangle) was identical to D = 2x_o + x_1.

This may be asking too much but can you find the mistake in my work? I just don't see it.

Thanks again.
 
  • #4
razored said:
Actually, I ignored the graph I drew when I solved the problem because I realized(on another piece of paper) that the area I was calculating (two triangles and rectangle) was identical to D = 2x_o + x_1.

This may be asking too much but can you find the mistake in my work? I just don't see it.

Thanks again.

Now that you know the answer, you can work backwards to figure the error in your math. In all honesty, I'm not that interested in looking.
 
  • #5
EDIT : My mistake.
 
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1. What is Iridov and how does it relate to kinematics?

Iridov is a unit used in kinematics to measure time intervals. It represents a time interval of 1.3 seconds.

2. How do you find the time interval using Iridov?

To find the time interval using Iridov, you simply multiply the number of Iridov by 1.3 seconds. For example, 2 Iridov would be equal to 2.6 seconds.

3. Can Iridov be used to measure shorter or longer time intervals?

Yes, Iridov can be used to measure both shorter and longer time intervals. It is a versatile unit that can be easily converted to fit different time scales.

4. What are some real-life applications of using Iridov to measure time intervals?

Iridov is commonly used in physics experiments and calculations, such as measuring the time it takes for an object to fall or the time it takes for a projectile to reach a certain distance. It is also used in sports timing, such as timing a runner's sprint or a swimmer's lap time.

5. How is Iridov different from other units used for measuring time intervals?

Iridov is unique in that it is specifically used in kinematics and is based on a decimal system, making it easy to convert to other units. Other units, such as seconds or minutes, may be more commonly used in everyday life but may not be as precise for scientific calculations.

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