V(t) = Vo + a t , Vo 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.