An electron moving with a velocity of 5 × 104 m/s enters into a uniform electric field and acquires a uniform acceleration of 104 ms-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?

Answer:

Given

Initial velocity (u) = 5 × 104 m/s

Acceleration (a) = 104 ms-2

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

Final velocity (v) = 2 u = 2 × 5 ×104 m/s

v = 10 × 104 m/s

To find t, use

v = at or t = u – u / a = (5 × 104)/104

t = 5s

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

Using s = ut + 1/2 at2

s = (5 × 104) × 5 + 1/2 × 104 × (5)2

s = 25 × 104 + (25 × 104) /2

s = 37.5×104 m

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