Question

Derive the equations:-


(i) v=u+at
(ii) S=ut+1/2at²

(iii) 2as=v²-u²​

Answer 1

Answer:

This is already there in your text book find it in your chapters

Answer 2

★Question :

Derive the equations:-

(i) v=u+at

(ii) S=ut+1/2at²

(iii) 2as=v²-u²

★⦋Answer⦌:

$(i) Consider \: \: a \: \: body \: \: having \: \: initial \: \: velocity \: \: 'u'. \: \: Suppose \: \: it \: \: is \: \: subjected \: \: to \: \: a \: \: uniform \: \: acceleration \: \: 'a' \: \: so \: \: that \: \: after \: \: time \: \: 't' \: \: its \: \: final \: \: velocity \: \: becomes \: \: 'v'. Now, \: \: from the \: \: definition \: \: of \: \: acceleration \: \: we \: \: know \: \: that: </p><p></p><p>Acceleration =Time \: \:  taken \: \: Change  \: \: in  \: \: velocity \: \: </p><p></p><p> \: or Acceleration = \: \: time \: \:  taken \: \: Final  \: \: velocity- Initial  \: \: velocity</p><p> \: </p><p> \: \: So, a=tv−u</p><p></p><p> \: \: at=v−u</p><p> \: and, v=u+at  \: \: </p><p>where  \: \: v=  \: final \: \: velocity \: \: of \: \: the \: \: body \: \: </p><p>u= initial \: \: velocity \: \: of \: \: the \: \: body \: \: </p><p> \: \: a= acceleration</p><p> \: \: and \: \:  t= time \: \: taken  \: \: </p><p></p><p> \: (b) Initial \: \: velocity,  \: \: u=54km/ \: \: h=15m/s \: \: </p><p>Final \: \: velocity, \: \:  v=0m/s \: \: </p><p>Time, t=8s</p><p> \: \: Acceleration, a=?</p><p></p><p> \: \: a=tv−u=80−15=8−15m/s2=−1.875m/s2 </p><p></p><p>$

(ii) There are in fact just TWO dynamics equations

Acceleration = Velocity Change / Time

Or: a=(v-u)/t more usually written as v=u+at

and Distance = Average Speed X Time

s = t (u+v)/2 usually written as that.

If you substitute for v from Eq.1 into Eq.2 you get

s = t(u + u+at)/2

which simplifies s = t (2u + at)/2

and goes further to s = ut + at²/2

The other suvat equations are derived by substitution in the same way, giving

v² = u² + 2as

s = vt - 1/2 at²

You could also sketch the v-t graph (a trapezium) and work out its area (a rectangle plus a triangle) and get

s = ut [the rectangle, base=t and height=u] + 1/2 t at [the triangle base=t and height=at]

We know that,

dv⃗ dt=a⃗

Multiplying and dividing LHS by ds⃗

$⟹dv⃗ \: ds⃗ \: ds⃗ \: dt=a⃗ </p><p>$

We replace ds⃗ dt with v⃗ to get

v⃗ dv⃗ ds⃗ =a⃗

⟹v⃗ dv=a⃗ ds

We know that at S=0,V=u and at S=s,V=v ,

Integrating both sides with appropriate limits:

⟹∫vuvdv=∫s0ads

⟹v22|vu=as|s0

⟹v22−u22=a(s−0)

⟹v2−u22=as

⟹v2−u2=2as

⟹v2=u2+2as

Or more appropriately,

v⃗ ⋅v⃗ =u⃗ ⋅u⃗ +2a⃗ ⋅s