Question

A mason is short with a constant speed of 7.5 107 m/s into a region, when an electric field
produces acceleration on the mason of magnitude 1.5 * 1015 ms-2 directed opposite to the
velocity. Find (a) the time taken (b) the distance covered by the mason before coming to rest and
(c) the time for which it remains at rest.​

Answer 1

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.  

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