Question

An electron moving with a velocity of
5x10^14 ms^-1 enters into a uniform electric field and
acquires a uniform acceleration of 10^4 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 1

a) Initial velocity of the electron is u = 5 x 104 m/s Acceleration of the electron is a = 104 m/s2

From first equation of motion,

v = u+ at

2u = u+ at

t=u/a = (5 x 104)/(104) = 5s

b) S = ut + 1/2 at^2

=> S = (5 x 1O^4) x 5 + 1/ 2 x 10^4 x 5^2

=> S = 3,75,000 m

Answer 2

Answer:

(i) Initial velocity of the electron is u = 5 × 104 m/sec

acceleration of the electron is a = 104 m/sec2

From (i) eqn of motion

v = u + at

2u = u + at

t = u/a = (5 × 104) = 5sec

(ii) S = ut + 1/2 at^2

Or, S = (5 × 10^4) × 5 + 1/2 × 10^4 × 5^2

Hence, S = 3,75,000 m

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