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An electron moving with a velocity of 5 × 104 m s-1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s–2 in the direction of its initial motion. 

(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. 

(ii) How much distance the electron would cover in this time?

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1 Answer

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Given initial velocity, u = 5 × 104 m s-1 

and acceleration, a = 104 m s–2 

(i) final velocity = v = 2 u = 2 × 5 ×104 m s–1 =10 × 104 m s–1 

To find t,      use  v  =  at 

                    or t = u - u / a

(ii)           Using s = ut + 1/ 2 at 2 

= (5 ×104) × 5 + 1/ 2 (10 ) × (5) 2 = 25 ×104 + 25 /2 ×104 = 37.5×104 m

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