Can anyone help with a Calculus proof?

Question

Using derivatives a = dv/dt, v = ds/dt and |v| = speed, show that if an object is moving on a straight line, then its speed is increasing when its velocity v and its acceleration a have the same sign and its speed is decreasing when v and a have opposite sign.

Answer 1

Jaina,

 

V(t) = V_o + a t , V_o is velocity at time t=0  along the straight line of motion.

V'(t) = a , the derivative of the velocity is acceleration along the straight line.

These equations show that V(t) is increasing if a is positive and decreasing if a is negative.

 

Assume the straight line path is inclined from the x-axis by an angle θ.  Then the velocity vector has x & y components given by:

Vx(t) = (Vo + at) cos (Θ)  ,  and Vy(t) = (Vo + at) sin (Θ) ,  where Vo and (Θ) are constants

Now  Vx²(t) + Vy²(t) =  (Vo + at)² [cos²(Θ) + sin²(Θ)] = (Vo + at)²  because cos² + sin² = 1

So  √[Vx²(t) + Vy²(t)]  = √[ (Vo + at)²] = (Vo + at)

 

But S(t) = | V(t) | = √[Vx²(t) + Vy²(t)]  = (Vo + at)

 

So S(t) also increase if a is positive and decrease if a is negative, same as V(t).  Proof is complete.