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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 ?

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 ? 

Grade:12

3 Answers

Khimraj
3007 Points
6 years ago
The distance covered by the meson in the electric field is given by 3rd equation of motion.
v2 = u2+2as
0 = 25*1012 -2*1.25*1014s
So s = 1/10m = 10cm
Rashmi
13 Points
2 years ago
Given,
Initial velocity (u)=5.0×106m/s
Acceleration (a) =1.25×104m/s^2
Final velocity (v) = 0
Then,
According to the third equation of motion,
V^2=u^2+2as
0=(5×10^6)^2 +2×1.25×10^4×s
s=1/10m
Or
S=10 cm
 
Sudhin Varghese
28 Points
2 years ago
Given that
The acceleration of meson = 1.25 × 10 14 m/s2
The initial velocity (u) = 5.0×106m/s
The final velocity (v) = 0
We know that the third equation of motion states that,
V2=u2+2as
Putting values,
0=(5×106)2 + (2×1.25×104×s)
0 = (25×1012) + (2.5×1014 × s)
s=1/10 m or 10 cm

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