The Speed Of a particle moving in a plane is equal to the magnitude of its instantaneous velocity, `v=|vl= sqrt(v_x^2+v_y^2).`
(a) Show that the rate of change of the speed is `(dv)/(dt)=(v_xa_x+v_ya_y)/(sqrt(v_x^2+v_y^2))`.
(b) Show that the rate of change of speed can be expressed as `(dv)/(dt)` is equal to`a_t` the component of a that is parallelto v.

Answer

A

Explanation

(b)`v=v_x hati+v_y hatj and a=a_x hati+a_y hatj.`
`v.a=v_xa_x+v_ya_y`
Further `v=sqrt(v_x^2+v_y^2)`
`:. (v.a)/v=(v_xa_x+v_ya_y)/(sqrt(v_x^2+v_y^2))`
`=(dv)/(dt)=a_1`
or component of a parallel to v
=tangential acceleration.