electron moving with a velocity of 5*10000 m/s enters into a uniform electric feild and acquires a uniform aceleration of 10000 m/s2 in direction of its initial motion
(1) calculate the time in which the electron would acquire a velocity double of its initial velocity
(2) how much distance electron would cover in this time

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Solution

(1)
Initial velocity, u = 50000 m/s
Acceleration, a = 10000 m/s²
Using,
v = u + at
=> 2u = u + at
=> t = u/a = 50000/10000
=> t = 5 s
In 5 s the electron velocity doubles the initial velocity.
(2)
Using,
S = ut + ½ at²
=> S = (50000)(5) + ½ (10000)(5²)
=> S = 375000 m
This is the distance traveled in that 5 s.

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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. 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.

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. 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. Question 22(ii)
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}

n the direction of its initial motion.

(ii) How much distance the electron would cover in this time?

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(1)Initial velocity, u = 50000 m/sAcceleration, a = 10000 m/s2Using,v = u + at=> 2u = u + at=> t = u/a = 50000/10000=> t = 5 sIn 5 s the electron velocity doubles the initial velocity.(2)Using,S = ut + ½ at2=> S = (50000)(5) + ½ (10000)(52)=> S = 375000 mThis is the distance traveled in that 5 s.