A bullet leaves a rifle with a muzzle velocity of 1042 m/s. While accelerating through the barrel of therifle, the bullet moves a distance of 1680 m. Determine the acceleration of the bullet (assume a uniform acceleration)
Answers
Acceleration of the bullet is a=323144.048\ m/s^2.
Explanation :
It is given that,
Initial velocity of the bullet, u = 0
Final velocity of the bullet, v = 1042 m/s
Distance covered by the bullet, s = 1.680 m
Now using third equation of motion :
v^2-u^2=2as
(1042\ m/s)^2-0=2a\times 1.680
a=323144.048\ m/s^2
Hence, this is the required solution.
Given that, a bullet leaves a rifle with a muzzle velocity of 1042 m/s.
(Initial velocity i.e. u of the bullet is 0 m/s and final velocity i.e. v of the bullet is 1042 m/s.)
Also, while accelerating through the barrel of therifle, the bullet moves a distance of 1680 m. (s = 1680 m)
We have to find the acceleration of the bullet.
Using the Third Equation Of Motion,
v² - u² = 2as
From the above data we have, u = 0/s, v = 1042 m/s and s = 1680.
Substitute the known values in the above formula,
→ (1042)² - (0)² = 2 × a × 1680
→ 1085764 - 0 = 3360a
→ 1085764 = 3360a
Divide by 3360 on both sides,
→ 1085764/3360 = 3360a/3360
→ 323.14 = a
→ a = 323.14
Therefore, the acceleration of the bullet is 323.14 m/s².