Question #2e2cc

Kinematics is a branch of mechanics used to describe motion—without regard to what caused that motion (forces). Kinematic equations are used to mathematically describe the motion of an object (e.g. displacement, velocity, acceleration).

Consider an object whose acceleration `a_s` remains constant during the time interval `Deltat=t_f-t_i`.

At the beginning of this interval, `t_i`, the object has initial velocity `v_(is)` and initial position `s_i`.

Say we want to predict the object's final position `s_f` and final velocity `v_(fs)` at time `t_f`.

The object's velocity is changing because the object is accelerating . We can find the object's velocity `v_(ts)` at a time `t_f`.

By definition:

`a_s=(Deltav_s)/(Deltat)=(v_(ts)-v_(is))/(Deltat)`

which is easily rearranged to give

`color(blue)(v_(fs)=v_(is)+a_sDeltat)`

The velocity-versus time graph is a straight line that starts at `v_(is)` and has a slope of `a_s`.

The object's final position is

`s_f=s_i+"area under the curve " v_s " between " t_i " and " t_f`

The area under the curve can be subdivided into a rectangle of area `v_(is)Deltat` and a triangle of area `1/2(a_sDeltat)(Deltat)=1/2a_s(Deltat)^2`.

Adding these gives:

`color(blue)(s_f=s_i+v_(is)Deltat+1/2a(Deltat)^2)`

where `Deltat=t_f-t_i` is the elapsed time

The final kinematic equation for constant acceleration is first found by using our first equation (in blue) to write:

`Deltat=(v_(fs)-v_(is))/a_x`

Substituting that into our second (blue) equation, we get:

`s_f=s_i+v_(is)((v_(fs)-v_(is))/a_x)+1/2a_s((v_(fs)-v_(is))/a_x)^2`

With a bit of algebra, this is rearranged to read

`color(blue)((v_(fs))^2=(v_(is))^2+2a_sDeltas)`

where `Deltas=s_f-s_i` is the displacement

Hope that helps!