An electron moving with a velocity of 5 × 10⁴ m s^-1 enters into a uniform electric field and acquires a uniform acceleration of 10⁴ m s^–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?

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Solution

Initial velocity of the electron is, u = 5 Ã— 10⁴ m/s
Acceleration of the electron is, a = 10⁴ m/s²
(i)
Using, v = u + at
=> 2u = u + at
=> t = u/a = (5 Ã— 10⁴)/( 10⁴) = 5 s
(ii)
Distance travelled in this time is,
S = ut + ½ at²
=> S = (5 Ã— 10⁴) Ã— 5 + ½ Ã— 10⁴ Ã— 5²
=> S = 375000 m

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An electron moving with a velocity of 5 × 104 m per second enters into a uniform electric feild and acquires a uniform acceleration of 104 m per second square in the direction of its initial motion

Calculate the time in which the electron would acquire a velocity double of its initial velocity and distance covered

Q. Question 22(i)
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Q. Question 22(i)
An electron moving with a velocity of

5 \times 10^{4} m s^{- 1}

enters into a uniform electric field and acquires a uniform acceleration of

10^{4} m s^{- 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.

Q. An electron moving with a velocity of

5 \times 10^{4} m s^{- 1}

enters into a uniform electric field and acquires a uniform acceleration of

10^{4} m s^{- 2}

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

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Initial velocity of the electron is, u = 5 Ã— 104 m/sAcceleration of the electron is, a = 104 m/s2(i)Using, v = u + at=> 2u = u + at=> t = u/a = (5 Ã— 104)/( 104) = 5 s(ii)Distance travelled in this time is,S = ut + ½ at2=> S = (5 Ã— 104) Ã— 5 + ½ Ã— 104 Ã— 52=> S = 375000 m