How to calculate distance from non-uniformly accelerated motion?

They gave a formula in my book for calculating the distance on a graph where the Y axis is of velocity motion, its to take the average start and end velocity and multiply it by the time. But how would you do that for non uniformly accelerated motion.How to calculate distance from non-uniformly accelerated motion?integration

it's all calculus



bit hard to explain trough this.tel me if u like i can explain



Kaushalya Damitha How to calculate distance from non-uniformly accelerated motion?If the curve is a straight line, then its slope everywhere is a = dv/dt = constant acceleration; where dv = v1 - v0 and dt = t1 - t0. In similar vein, dS = s1 - s0 = v dt = ((v1 + v0)/2) (t1 - t0); where dS is the distance traveled in the interval t1 - t0. Here v1 %26gt; v0, t1 %26gt; t0, and average velocity v = (v1 + v0)/2 We can simply divide the sum of the two velocities by 2 for the average velocity as the slope of the curve, a = dv/dt, is a fixed value.



But when the graph line is curved dv/dt %26lt;%26gt; constant, but the acceleration does depend on where the slope of the curve is meaured. And that depends on the independent variable, which is time. Thus we write a(t) = dv/dt, which means a depends on t. As a depends on t, v also depends on t because velocity springs from the acceleration. Then, finally, S(t) means that the distance traveled also depends on the time since time dependent acceleration determines time dependent velocity and time dependent velocity determines time dependent distance traveled.



Bottom line, you need to know the formula for the v(t) = f(t) relationship, which is what determines the shape of the line on your v-t graph. f(t) is math talk for "a function of t". So the equation state velocity at time t is a function of t.



Let's look at an example. Suppose v(t) = v0 + t^2 and v(0) = 0 is the function that determines the shape of the line in your graph. That is, at time zero, the velocity is zero. Or another way to put it, the line will intercept the Y axis where the X axis = 0 on your v-t graph.



Then dv(t)/dt = a(t) = 2t is the acceleration at any time t on the line. The time dependent distance is dS(t)/dt = v(t) = v0 + t^2; so that dS(t) = v0 dt + t^2 dt. And we can find S = INT(dS) = INT(v0 dt + t^2 dt) = v0 t + (t^3)/3 would be the distance traveled in the time interval t. INT stands for integral over the time interval t = t1 - t0.



Bottom line, if the line is straight, getting the distance traveled is easy because acceleration is uniform (fixed) and we can use v = (v1 + v0)/2 in S = vt to find the distance traveled. But if the line is curved, you need to know the equation for the line in terms of velocity v(t) and time. Then you need to use differential and/or integral calculus to find the distance traveled S(t).



OR, and this is a big OR, find the area under the curve in your v-t graph. Since S(t) = SUM(v(t) dt), the area under the curve between t1 and t0 will yield the distance traveled no matter if the line is curved or straight.