Question

A meson is shot with constant speed 5.0 × 106 m/s in an electric field which produces

on the meson an acceleration of 1.25 × 1014 m/s2 directed opposite to the initial

velocity. How far does the meson travel before coming to the rest ?​

Answer 1

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Answer:

Distance, d = 0.1 m

Explanation:

It is given that,

Initial velocity of meson,

Finally, the meson is coming to rest v = 0

Acceleration of the meson, (opposite to initial velocity)

Using third equation of motion as :

s is the distance the meson travelled before coming to rest.

So,

 

s = 0.1 m

The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.