Russ P. answered • 02/09/15

Tutor

4.9
(135)
Patient MIT Grad For Math and Science Tutoring

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 Vx

^{2}(t) + Vy^{2}(t) = (Vo + at)^{2}[cos^{2}(Θ) + sin^{2}(Θ)] = (Vo + at)^{2}because cos^{2}+ sin^{2}= 1So √[Vx

^{2}(t) + Vy^{2}(t)] = √[ (Vo + at)^{2}] = (Vo + at)But S(t) = | V(t) | = √[Vx

^{2}(t) + Vy^{2}(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.