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. How much distance would the electron cover in this time?
ScienceClass 9CBSEAnswered 27 Mar 2026
Answer
Given initial velocity, u = 5 × 104 m/s and acceleration, a = 104ms-2
- final velocity = v = 2 u = 2 × 5 ×104 m/s =10 × 104 m/s To find t, use v = at or t = u – u / a = (5 × 104)/104
=5s
- Using s = ut + 12
at 2 = (5 ×104) × 5 +
12
(10 ) × (5) 2
= 25 ×104 + 25 /2 ×104
= 37.5×104 m
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