How long does it take for a car to travel 100 meters with a given acceleration?

In summary, the conversation discusses how to find the time it takes for a car to travel 100 meters given its acceleration and starting velocity. The solution involves integrating the acceleration to find the velocity and then integrating again to find the position. The final step is to set the position equation equal to 100 and solve for time. The conversation also mentions the importance of choosing an appropriate reference point for measuring distances.
  • #1
ada0713
45
0

Homework Statement



A car starts from rest at time t=0 and accelerates at -0.6t+4 meters/sec^2
for 0[tex]\leq[/tex]t[tex]\leq[/tex]12. How long does it talke for the car to go 100 meters?

The Attempt at a Solution


Since it says the acceleration is -0.6t+4
I integrated it and ended up with v(t)=-0.3t^2 + 4t + C for the velocity
but I don't know what to do after this.
I tried to intergrate it agian so that I can find the distance equation and set that
equal to 100 but there is an unkown constant C that I have to figure out first.
HELP!
 
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  • #2
okay. Since it syas the car starts from the rest
v(0)=-0.3(0)^2+4(0)+c=0
therefore v(t)=-0.3t^2+4t
I got this far.
But when I integrate this,
I get s(t)=-0.1t^3+2t^2+C
another constant...
Now what do I do?
 
  • #3
Again, you fill in an initial condition. For example, you can call the position of the car at t = 0 the origin, so s = 0. Then the question, when the car has traveled 100 m, comes down to solving s = 100.
You can also leave the constant in. Then the question is to solve
[tex]s(t) - s(0) = 100[/tex]
for t. Though this is slightly more complicated, you can try it: you'll see that the constant drops out in the end, that is: the requested time is not dependent on the initial point with respect to some special reference point, just on the time difference with respect to the (arbitrary) starting point.

Basically, it's an arbitrary choice: you have to choose which point you call s = 0, that is, relative to which point you're going to measure your distances... might as well be a logical point. Compare it to this situation: I give you a ruler and ask you to measure the width of your screen. You might put the zero of the ruler on the left side, and read off the length from the right side directly. Or you might just hold the ruler up to the screen, read off the left hand side and right hand side marks, and subtract them. In both cases you will get the correct answer. It's just a matter of which point you call zero, but in the end it's about the difference of the number that is on the right and left side. It just makes life easier -- if you have the choice anyway -- to make the left side be zero (by shifting your ruler, or your function s(t)) such that in the end, you will be subtracting zero instead of some arbitrary number C, to get the difference you want.
 
Last edited:
  • #4
That means s(0)=0 right? then I got 0 for constant again
s(t)=-0.1t^3+2t^2
so I set
-0.1t^3+2t^2=100

t^3-20t+1000=0
now solve for t?
it seems the answers going to be pretty complicated,,
Am I doing it right?
 

1. What are construction antiderivatives?

Construction antiderivatives are mathematical tools used to help calculate the area underneath a curve in a graph. They are the inverse operation of derivatives, which are used to find the slope of a curve at a specific point. Construction antiderivatives are also known as indefinite integrals.

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In construction, antiderivatives are used to calculate quantities such as the volume of concrete needed for a specific shape or the amount of materials required for a particular project. They are also used to determine the area of irregularly shaped land or to find the total distance traveled by a construction vehicle.

3. What is the process for finding a construction antiderivative?

The process for finding a construction antiderivative involves using the fundamental theorem of calculus, which states that the antiderivative of a function can be found by integrating the original function. This involves reversing the steps of taking a derivative, such as applying the power rule or product rule, to find the original function.

4. Can construction antiderivatives be used to solve real-world problems?

Yes, construction antiderivatives are commonly used in real-world applications, particularly in engineering and construction. They can be used to calculate quantities and solve problems related to area, volume, distance, and other physical properties.

5. Are there any limitations to using construction antiderivatives?

While construction antiderivatives can be powerful tools for solving problems, there are some limitations to their use. For example, they may not always provide an exact solution and may require the use of estimation techniques. Additionally, they may not be applicable to all types of functions, such as those with discontinuities or infinite intervals.

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