Question

A mason is shot with a constant speed of 7.5 x 10²m/sinto a region, when an electric field produces acceleration on the mason of magnitude 1.5 x 1025 ms-2 directed opposite to the velocity. Find 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

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.