Integrating Kinematics for Velocity from Acceleration: A Simplified Approach

yup790
Messages
21
Reaction score
0
When you want to get velocity from accelleration i have been told you integrate.

Howver v=at and so surley you can just multiply each term in the accelleratin expression by t.

ie:
a=4-0.2t

Surley you can just:
v=(4-0.2t)t
v=4t-0.2t2
 
Physics news on Phys.org
The equation v=at is for the situation when the acceleration is a constant. If it is a function of t, you have to integrate. In that case the corresponding equation is dv=a(t)dt, which gives the infinitesimal change in velocity, dv during infinitesimal time interval dt when the acceleration function a(t) is known. When integrating over a finite time interval, you effectively add a large number of small velocity changes dv to get the total change in velocity, Δv.
 
Last edited:
Thank you. Is there any proof for this. I learn better when I understand the theory behind a topic.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top