Define (i) v = u + at (ii) V2 – u2 = 2as by calculus method
CLASS - XI PHYSICS (Kinematics)
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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
\int\limits^u_v \, dv=a \int\limits^0_t \, dt
[v]^v_u=a[t]^t_0
v-u=at
v=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
\int\limits^v_u {v} \, dv =
[\frac{v^2}{2}]^v_u =a[x]^s_0
v^{2}-u^{2} = 2as
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
\int\limits^u_v \, dv=a \int\limits^0_t \, dt
[v]^v_u=a[t]^t_0
v-u=at
v=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
\int\limits^v_u {v} \, dv =
[\frac{v^2}{2}]^v_u =a[x]^s_0
v^{2}-u^{2} = 2as
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