question_answer

An electron moving with a velocity of x 104 m/s enters into a uniform electric field and acquires a uniform accelerations of 104 m/s2 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:

                       Given, u = 5 X 104 m/s, a = 104 m/s2 Final velocity, \[\upsilon =2u=2\times 5\times {{10}^{4}}={{10}^{5}}\text{m/s}\] \[\upsilon =u+at\]                 \[\Rightarrow \] \[t=\frac{\upsilon -u}{a}=\frac{{{10}^{5}}-5\times {{10}^{4}}}{{{10}^{4}}}\]                 \[=\frac{{{10}^{4}}(10-5)}{{{10}^{4}}}=5s\]                                     (ii) Distance \[s=ut+\frac{1}{2}a{{t}^{2}}\]                            \[=5\times {{10}^{4}}\times 5+\frac{1}{2}\times {{10}^{4}}\times {{(5)}^{2}}\] \[=25\times {{10}^{4}}+\frac{25}{2}\times {{10}^{4}}\] \[=37.5\times {{10}^{4}}m\]