CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An electron moving with a velocity of 5×104 ms1 enters into a uniform electric field and acquires a uniform acceleration of 104 ms2 in the direction of its initial motion. How much distance the electron would cover in 5 s?

A
37.5×103 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
37.5×104 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
30.5×104 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
30.5×105 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 37.5×104 m
Given:
initial velocity, u=5×104 ms1,
acceleration, a=104 ms2, and
t=5 s

Using s=ut+12at2
=(5×104)×5+12(104)×(5)2=25×104+252×104=37.5×104m
= 37.5×104 m

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Equation of Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon