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Question
An electron moving with a velocity of 5 × 104 ms−1 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 at this time?
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Solution
Here initial velocity of the electron,
u = 5 × 104 ms−1
acceleration, a = 104 ms−2
(i) v = 2u = 2 × 5 × 104 ms−1
= 10 ×104 ms
using v = u + at we get
10 × 104 = 5 × 104 + 104 × t or t = 5s
(ii) Using s = `"ut" + 1/2 "at"^2`, we get
s = `5 xx 10^4 xx 5 + 1/2 xx 10^4 xx 5^2`
= 25 × 104 + 12.5 × 104
= 37.5 × 104 m.
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