Question

An electron moving with the speed of 5 x 10^4 m/s enters into an electric field and attains a uniform acceleration of 10^15 m/s^2 in the direction of motion . In how much time will it attain a speed twice of its initial speed ? In this time , how much distance will it cover ?

Answer 1

Answer: u=5*10m/s

a=10^15m/s²

v=2u=2*5*10⁴m/s=10⁵m/s

Applying v=u+at

t=v-u/a=2u-u/a=u/a=5*10⁴m/s/10¹⁵m/s²=5*10^-11s

Applying s=it+1/2at²

=25*10^-7m+12.5*10^-7m

=37.5*10^_7=3.75*10^-6m

Answer 2

Answer:

Time= 5s

Explanation:

We have u=5 x 10^4 m/s

acceleration= 10^15 m/s

we need final velocity to be twice so

2u=u+at

u=at

t=u/a

t=5 x 10^4 m/s divided by 10^15 m/s

0.00000000005 s

Distance covered will be=

s=ut+1/2at^2

0.0000025+ 125*10^14