derive v²=u²+2as by calculas method.
Answers
Answer:
acceleration is taken as constant
we know a = dv/dt
dv = a dt
Integrating both sides with proper limits
\int\limits^u_v \, dv= \int\limits^0_t {a} \, dt
v
∫
u
dv=
t
∫
0
adt
\int\limits^u_v \, dv=a \int\limits^0_t \, dt
v
∫
u
dv=a
t
∫
0
dt
[v]^v_u=a[t]^t_0[v]
u
v
=a[t]
0
t
v-u=atv−u=at
v=u+atv=u+at
second equation
a= dv/dt x dx/dx
a = v dv/dx
v dv = a dx
Integrating both sides with proper limits
\int\limits^v_u {v} \, dv = \int\limits^s_0 {a} \, dx
u
∫
v
vdv=
0
∫
s
adx
\int\limits^v_u {v} \, dv =
u
∫
v
vdv=
[\frac{v^2}{2}]^v_u =a[x]^s_0[
2
v
2
]
u
v
=a[x]
0
s
v^{2}-u^{2} = 2asv
2
−u
2
=2as
Answer:
Explanation:
as we know rate of change of velocity is accleration
dV/dt = a
multiplying deominator and numerator with dx
dV/dt * dx/dx = a
re arranging , dV/dx * dx/dt = a
we know rate of change of distance = dx/dt = V
dV / dx * V = a
cross multiplying VdV = adx
intergrating on both sides
=
v