Question 22(ii)
An electron moving with a velocity of 5×104ms−1 enters into a uniform electric field and acquires a uniform acceleration of 104ms−2 n the direction of its initial motion.
(ii) How much distance the electron would cover in this time?
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Solution
Given initial velocity, u=5×104ms−1 and acceleration, a=104ms−2
(ii) Using s=ut+12at2 =(5×104)×5+12(104)×(5)2=25×104+252×104=37.5×104m