dstebbins Posted November 30, 2013 Share Posted November 30, 2013 I was looking for the equation for acceleration, and I found this webpage: http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienceIntegation-836.htm However, this equation doens't help me, because in my current situation, I don't know the final velocity. I know the starting velocity was 0 m/s, because the object whose acceleration I'm trying to calculate started in a stationary position. What's the equation for figuring out the acceleration if you begin at rest, and then traverse d meters over the course of t seconds? Obviously, the AVERAGE velocity would be d/t, but how do I calculate the acceleration? Link to comment Share on other sites More sharing options...

Enthalpy Posted November 30, 2013 Share Posted November 30, 2013 If the acceleration is contstant, the average velocity is half the final velocity. Link to comment Share on other sites More sharing options...

Endercreeper01 Posted November 30, 2013 Share Posted November 30, 2013 The acceleration would be the second derivative of the distance with respect to time, which is written as d^{2}s/dt^{2} Where s is distance and t is time Link to comment Share on other sites More sharing options...

dstebbins Posted November 30, 2013 Author Share Posted November 30, 2013 Ok, hypothetical scenario: If I'm in a parked car, and then I step on the gas, traveling the hypotnuse of a 100m x 100m right triangle in 10 seconds, then I'd be traveling a straight distance of 141.42 meters, right? So, if I do so in 10 seconds, then my average speed is 14.14 m/s, right? So, you're saying that my final velocity is 28.28 m/s? So, that would mean that my average acceleration would be 2.83 m/(s^2)? Is that about right? The acceleration would be the second derivative of the distance with respect to time, which is written as d^{2}s/dt^{2} Where s is distance and t is time So, what's the d stand for? Link to comment Share on other sites More sharing options...

Endercreeper01 Posted November 30, 2013 Share Posted November 30, 2013 d would be for the derivative. If you had a function f(x) which is the same as y, then you could write the derivative 2 ways. One is f'(x). The second is dy/dx. With the second derivative, you could also write it 2 ways. One is f''(x). The other is d^{2}y/dx^{2} Link to comment Share on other sites More sharing options...

studiot Posted November 30, 2013 Share Posted November 30, 2013 (edited) I think that you need to find a better website, that one was rather elementary. Science Forums is a good one If u is the initial velocity v is the final velocity s is the distance travelled t is the time is the time of travel f is the acceleration, which must be constant throughout t There are several equations to choose from when doing these problems. v = u +ft v^{2} = u^{2}+ 2fs s = ut + 1/2 ft^{2} s = vt - 1/2 ft^{2} s = 1/2(v+u)t Remember that deceleration is negative acceleration ie f is negative for deceleration. Edited November 30, 2013 by studiot Link to comment Share on other sites More sharing options...

swansont Posted December 1, 2013 Share Posted December 1, 2013 I was looking for the equation for acceleration, and I found this webpage: http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienceIntegation-836.htm However, this equation doens't help me, because in my current situation, I don't know the final velocity. I know the starting velocity was 0 m/s, because the object whose acceleration I'm trying to calculate started in a stationary position. What's the equation for figuring out the acceleration if you begin at rest, and then traverse d meters over the course of t seconds? Obviously, the AVERAGE velocity would be d/t, but how do I calculate the acceleration? d = 1/2 at^{2 }if the acceleration is constant and it starts at rest. Thus, a = 2d/t^{2} This is from the basic kinematics equation s = v_{0}t + 1/2 at^{2 }which you get by taking a = dv/dt integrating twice (also using v = ds/dt). If acceleration is not constant, then you don't have enough information to solve the problem. Link to comment Share on other sites More sharing options...

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