An electron moving with a velocity of 5×104ms-1 enters into a uniform electric field and acquires a uniform acceleration of 104ms-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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Step 1: GivenThe electron&#039;s initial velocity, u=5×104ms-1Acceleration acquired by the electron, a=104ms-2Step 2: Formulas usedThe three motion equations tell us that,Distance covered, s=ut+12at2,Where u represents the starting speed., t is the time taken and a is the acceleration of the body.We also know that,time, ∆t=v-ua∵a=∆v∆tStep 3: Calculate the time required to double the initial velocityWe need that the final velocity is twice the initial velocity, thus, v=2u=2×5×104=105ms-1.Thus, the time required to acquire double the initial velocity is, t=10×104-5×104104=5sStep 4: Calculate the distance covered in the meantimeBy using the first equation, the distance covered iss=5×104×5+12×104×52∴s=37.5×104mHence,It takes time to double the starting velocity of 5s.Distance covered while doubling the initial velocity is 37.5m.

An electron moving with a velocity of 5 xx 10^4 ms^(-1) enters into a uniform electric field and acquires a uniform acceleration of 10^4 ms^(-2) in the direction of its initial motion

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