Question

A positively charged meson is shot with a constant velocity of 7.5x10^7 m/s towards a positively charged plate which imparts it a constant acceleration of 1.25x10^15 m/s^2. Calculate the (i) time taken (ii) distance moved by the meson to come to rest. Also how long does it stay at rest?


Class XI______Motion in a straight line

Answer 1

Answer:

(i) 6 × $10^{-8}$ second.

(ii) 2.25 meter.

If you watch such an experiment at so tiny scale, as the radius of a meson is in -15 power of 10, there'll not be any moment when the meson stays at rest.

Explanation:

Since, v = u + at

0 = 7.5 × $10^{7}$ + ( -1.25 × $10^{15}$ ) × t

t = 6 × $10^{-8}$ second.

Also,

$v^{2} = u^{2} + 2as\\\\s= \frac{v^{2}-u^{2}}{2a}$

s = 2.25 meter.