Question

An electron moving with a velocity of 5×10^4 m/s enters into a uniform electric field and acquiers a uniform acceleration of 10^4 m/s^2 in the direction of its initial motion
1. Calculate the time in which the electron would acquire a velocity doubke of its initial velociry and
2) how much distance the electron would cover in this time

Answer 1

Given data:Initial velocity of the electron is,
u = 5 × 10^4 m/s

Acceleration of the electron is,
a = 10^4 m/s2

(i)

Using, v = u + at

=> 2u = u + at. [since v is double of Initial velocity (u)]

=> t = u/a = (5 × 10^4)/( 10^4) = 5 s

(ii)

Distance travelled in this time is,

S = ut + ½ at2

=> S = (5 × 10^4) × 5 + ½ × 10^4 × 5^2

=> S = 3,75,000 m

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Answer 2

Here ,u=5×10^4m/s

a = 10^4m/s

¡) from , v = u + at

2u = u + at ,

t =u/a = 5×10^4/10^4

=5s

¡¡) from s = ut + 1/2 at^2

s = 5×10^4×5 + 1/2 ×10^4 (5)^2

= 37.5×10^4m