allrightalready Posted May 23, 2015 Share Posted May 23, 2015 is this enough or is it the math relationship you need explained "This discussion illustrates that a free-falling object that is accelerating at a constant rate will cover different distances in each consecutive second. Further analysis of the first and last columns of the data above reveal that there is a square relationship between the total distance traveled and the time of travel for an object starting from rest and moving with a constant acceleration. The total distance traveled is directly proportional to the square of the time. As such, if an object travels for twice the time, it will cover four times (2^2) the distance; the total distance traveled after two seconds is four times the total distance traveled after one second. If an object travels for three times the time, then it will cover nine times (3^2) the distance; the distance traveled after three seconds is nine times the distance traveled after one second. Finally, if an object travels for four times the time, then it will cover 16 times (4^2) the distance; the distance traveled after four seconds is 16 times the distance traveled after one second. For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel." http://www.physicsclassroom.com/class/1DKin/Lesson-1/Acceleration Link to post Share on other sites
Trava u doma Posted May 23, 2015 Share Posted May 23, 2015 You can get this result by integration. But the easiest explanation is this: Assuming the initial speed is zero, and you accelerate with the acceleration a, the final speed will be a*t. Since the acceleration is linear, the mean speed will be a*t/2. Now, distance will be the time travelled times the mean speed, so (a*t/2)*t Does it make sense? And for reference, by integration you get this: Acceleration is the rate of change of speed, so a=dv/dt Speed is the rate of change of distance so v=ds/dt Therefore: v= adt = a*t + v_initial s= vdt= (a*t + v_initial)dt = (a*t^2)/2 + v_initial*t + s_initial Link to post Share on other sites
Asex Posted May 24, 2015 Share Posted May 24, 2015 The object's speed doesn't stay the same, it is constantly changing so we must come up with some sort of value that takes into consideration all the different speeds, from the first moment to the last moment of interest. What we are looking for is the mean speed. In the formula you gave the initial speed is zero, this is before the acceleration had a chance to make a difference. At the last moment the speed is the acceleration * time. In the formula you give the acceleration is constant, it means speed increases at a steady/fixed rate. So we draw a coordinate system, the horizontal line being for time, the vertical being for speed. We note the point of initial speed(and time), note the point of terminal speed and note the point of the last moment. We connect the three points and we must get a triangle like that (if the object is braking/decelerating it would be a mirror image) : /| / | / | / | /__| Now this triangle has an area and its area happens to equal the distance traveled by the object. The legs/catheti are time(horizontal leg) and speed(vertical leg) so we must multiply time * mean speed. By the way for displacement ( like a piston moving in a cylinder ) you must have some area * distance traveled so the difference of the volumes enclosed between the first and the last position of the area (subtract the smaller from the bigger) is the displacement. Link to post Share on other sites
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