Question

A meson is shot with constant speed 5.0 × 10 6 m/s in an electric field which produces on the meson an acceleration of 1.25 × 10 14 m/s 2 directed opposite to the initial velocity. How far does the meson travel before coming to the rest ?

Answer 1

your answer is in above picture..hope ut helped..

Answer 2

Answer:

Distance, d = 0.1 m

Explanation:

It is given that,

Initial velocity of meson, $u=5\times 10^{6}\ m/s$

Finally, the meson is coming to rest v = 0

Acceleration of the meson, $a=-1.25\times 10^{14}\ m/s^2$ (opposite to initial velocity)

Using third equation of motion as :

$v^2-u^2=2as$

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

So, $s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(5\times 10^{6})^2}{2\times -1.25\times 10^{14}}$  

s = 0.1 m

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