Question

give the derivation of
the following with the help of calculus

Answer 1

s is displacement
v is initial velocity
t is time
a is acceleration

Answer 2

Basic Physics !

Answer 3

you don't need calculus (for constant acceleration). You can prove it with the help of a graph.

Answer 4

i have to prepare for exam .... and the question is to prove it using calculus

Answer 5

@apoorvk

Answer 6

@ParthKohli

Answer 7

@Diyadiya

Answer 8

@musicalrose

Answer 9

@biswajit_paul

Answer 10

@biswajit_paul plz answer

Answer 11

OK answering it

Answer 12

s = ∫ v dt

s = ∫ (at + u) dt

s = ½ at² + ut + c

s(0) = 0

0 = ½ a * 0² + u * 0 + c

c = 0

s = ½ at² + ut

Answer 13

Gimme the medal !

Answer 14

\[-a=d ^{2}s/dt ^{2}=d/dt(ds/dt)\](as accelaratio is negative)
\[d(ds/dt)=-a.dt\]
\[ds/dt=-a.t+c\](after integrating, and c is a integration constant)
as ds/dt is instantaneous velocity so when t=0 then ds/dt=v(according to your que v is initial velocity)=c
so,
\[ds/dt=v-at\]
\[ds=v.dt-a.t.dt\]
\[s=vt-(1/2)at^{2}+k\](after integrating, k is another integrating constant)
now as at t=0 no distance was traversed so at t=o

s=0=k

so finally we proved

\[s=vt-(1/2)at^{2}\]

Answer 15

thanks @biswajit_paul

Answer 16

I have assumed that the particle (or object) was under retarding motion (negative accelaration)

Answer 17

I have done this from the basic definition of acceleration and using calculus.

Answer 18

very very thanks @biswajit_paul
धन्यवाद

Answer 19

hm..no mention