CH. VII ACCELERATION 153 



the time factor. Take fo equal to the final speed. Draw 

 a vertical line through a, cutting a horizontal through 

 / in d. Join od. Then od is the tangent to the accelera- 

 tion curve at the origin, and dafao will be the initial 

 acceleration. 



We know that when the time from the start is equal to 

 the time factor the speed is 0-632 of the final speed. Take 

 ab equal to this fraction of ad. The speed at a will then 

 be represented by ab. 



We know also that the acceleration when t = r is equal 

 to the initial acceleration divided by e. Draw a line 

 through &, making an angle with the horizontal whose 

 tangent is this fraction of dajao, this will be the tangent 

 to the curve at />. 



In Fig. 38 curve B gives the current drawn from the 

 line. Since the current at any instant is given by 



c= , n being the speed at that instant, the equation 

 H 



to the current curve is 



(79). 



The area of this curve gives the work done, since the 



tension of the line remains constant. If ok is the final 



t 



current and op the initial current, then c = ok + kp x e r - 

 The tangent to the current curve at the origin is ~-. If we 



T 



take ag equal to ok, and join pg, this will be the tangent to 



kp 

 the current curve at p. The current at t = r is ok + -1- and 



e 



the tangent to the curve here is the initial value of the 

 tangent divided by e. 



