Dabour Hub EGEgyptian Curriculum 2026/2027
2-1 (a) Newton's second law as a differential equation

2-1 (a) Newton's second law as a differential equation

Point 1 of 7 — Motion with Variable Acceleration and ForceQuestion bank: 0 questions (0 with pictures)

01 2-1 (a) Newton's second law as a differential equation

The law

  • The rate of change of momentum is proportional to the force and in its direction: F⃗=ddt(mv⃗).
  • When the mass is constant this gives F⃗=ma⃗ with the units chosen so that the constant of proportionality is 1 (newton for m in kg and a in m/s2; dyne for g and cm/s2).
  • 1\ \text{kg·wt} = 9.8 N and 1 gf=980 dyne.

Three forms of the acceleration

  • Acceleration is a=dvdt, and by the chain rule with v=dsdt we also have a=dvdt=dvds⋅dsdt=v dvds=dds(12v2)
  • Which form to use depends on what the force depends on.
Point of comparison Force depends on time t Force depends on position s Force depends on velocity v
Equation used mdvdt=F(t) m vdvds=F(s) mdvdt=F(v) or m vdvds=F(v)
Variables separated dv=F(t)mdt v dv=F(s)mds dvF(v)=dtm or v dvF(v)=dsm
First answer v(t) v(s) (speed at a place) v(t) or v(s)
Then integrate again for s(t) dsdt=v(s) for t integrate for the other variable

Integrals that follow

  • Impulse–momentum: ∫t1t2F dt=m(v2−v1).
  • Work–energy: ∫s1s2F ds=12m(v22−v12).
  • Constant acceleration is a special case: a=c gives v=v0+ct, s=v0t+12ct2 and v2=v02+2cs.

General method

  • Draw the body and the forces; take the direction of motion as positive.
  • Write the equation of motion Fresultant=ma and choose the form of a.
  • Separate the variables and integrate between the initial and final values (limits), or integrate indefinitely and find the constant from the initial condition.
  • Keep units consistent.

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