Motion Equations Assignment Help

Classical Physics - Motion Equations

Motion Equations

(a) V = u + at; (b) s = ut + 1/2 at2

(c) v– u2 = 2as; (d) snth = u + a/2 (2n – 1)

Derivations

(a) A = dv/dt or dv = a dt

∫vu dv = ∫to adt

V – u = at or v = u + at

(b)    V = ds/  dt or ds = v .dt = ( u + at) dt

∫so ds = ∫to (u + at) dt

S = ut + at2 / 2 


(c) Eliminate t from equations (I) and (2)

T= v –u / a or s = u (v – u / a) + a/ 2 (v – u / a)2

Or v– u2 = 2as

(d) Sn = un + a / 2 n
2

Sn-1 = u (n – 1) = a/ 2 (n – 1)2

Snth = (Sn – Sn -1) = u + a/2 (2n – 1)


The conditions under which these equations can be applied

(a) Motion should be one dimensional 

(b) 
Acceleration should be uniform

(c) Frame of reference should be inertial

Procedure to draw graphs

Consider the equation v = u + at

Compare the above equation with y = c + ms,

If v is along y-axis, t along x axis, then it is an equation of a straight line.

If c - 0, that is, y = mx or v = at (u = 0)

Then it is the equation of a straight line passing through origin. 

Consider s = ut + 1/2 at2

Compare the above equation with y = ax + bx2. If s is (taken along y – axis t taken along x – axis) which is the general equation of a parabola.

Now even if a = 0, that is, y = bx2 or s = i/2 at2

It is still the equation of a parabola. However if acceleration is zero that is, b = 0 in general equation theny = ax or s = ut which is the equation of a straight line (passing through origin)

While drawing the curves these equations are to be kept in mind and accordingly the graph is drawn.

The other important equations are 

x2 + y2 = a2 (equation of circle)

x2/a2 + y2/b2 = 1 (equation of an ellipse)

x2/a2 – y2/b2 = 1 (equation of hyperbola)

xy = c2 -> (equation of hyperbola)

Y = e-ax (exponential)


For example, radioactive disintegration or discharging of a capacitor and so on. 

Y = y0e-bx 

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