enters a region with a speed of 530 times 106 ms and then is

enters a region with a speed of 5.30 times 10^6 m/s and then is slowed at the rate of 2.80 times 10^14 m/s^2. How far does the muon take to top?

Solution

Here,

u=5.30x10^6 m/s
v=o
a= - 2.80 x10^14m/(s^2).
s=?
Use the equation of motion : v^2= u^2+2as

put the values -

0 = (5.30x10^6)^2 - 2*2.80 x10^14*s

=> s = (28.09x10^-2) / 5.6 = 5.02 x 10^-2 m = 5.02 cm