Question

– is second kinematics equations​

Answer 1

Answer:

Example 2: Second kinematic formula, Δ x = ( v + v 0 2 ) t {\Delta x}=(\dfrac{v+v_0}{2})t Δx=(2v+v0)t.

Explanation:

Answer 2

The acceleration of a particle is the vector defined by the rate of change of the velocity vector. The average acceleration of a particle over a time interval is defined as the ratio.

${\displaystyle {\overline {\mathbf {A} }}={\frac {\Delta \mathbf {V} }{\Delta t}}\ ,}{\overline {\mathbf {A} }}={\frac {\Delta \mathbf {V} }{\Delta t}}\ ,$

where ΔV is the difference in the velocity vector and Δt is the time interval.

The acceleration of the particle is the limit of the average acceleration as the time interval approaches zero, which is the time derivative,

${\displaystyle \mathbf {A} =\lim _{\Delta t\rightarrow 0}{\frac {\Delta \mathbf {V} }{\Delta t}}={\frac {d\mathbf {V} }{dt}}={\dot {\mathbf {V} }}={\dot {v}}_{x}{\hat {\imath }}+{\dot {v}}_{y}{\hat {\jmath }}+{\dot {v}}_{z}{\hat {k}}}{\displaystyle \mathbf {A} =\lim _{\Delta t\rightarrow 0}{\frac {\Delta \mathbf {V} }{\Delta t}}={\frac {d\mathbf {V} }{dt}}={\dot {\mathbf {V} }}={\dot {v}}_{x}{\hat {\imath }}+{\dot {v}}_{y}{\hat {\jmath }}+{\dot {v}}_{z}{\hat {k}}}$

or

${\displaystyle \mathbf {A} ={\ddot {\mathbf {P} }}={\ddot {x}}_{p}{\hat {\imath }}+{\ddot {y}}_{P}{\hat {\jmath }}+{\ddot {z}}_{P}{\hat {k}}}{\displaystyle \mathbf {A} ={\ddot {\mathbf {P} }}={\ddot {x}}_{p}{\hat {\imath }}+{\ddot {y}}_{P}{\hat {\jmath }}+{\ddot {z}}_{P}{\hat {k}}}$

Thus, acceleration is the first derivative of the velocity vector and the second derivative of the position vector of that particle. Note that in a non-rotating frame of reference, the derivatives of the coordinate directions are not considered as their directions and magnitudes are constants.

The magnitude of the acceleration of an object is the magnitude |A| of its acceleration vector. It is a scalar quantity:

${\displaystyle |\mathbf {A} |=|{\dot {\mathbf {V} }}|={\frac {dv}{dt}},}{\displaystyle |\mathbf {A} |=|{\dot {\mathbf {V} }}|={\frac {dv}{dt}},}$