22 PE
Question
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of \({\bf{6}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) for\({\bf{8}}{\bf{.10 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{s}}\) . What is its muzzle velocity (that is, its final velocity)?
Step-by-Step Solution
VerifiedThe muzzle velocity of bullet is \(502.2\;{\rm{m/s}}\).
Given Data:
The initial velocity of bullet is \(u = 0\)
The time for acceleration of bullet is \(t = 8.10 \times {10^{ - 4}}\;{\rm{s}}\)
The acceleration for bullet is \(a = 6.20 \times {10^5}\;{\rm{m}}/{{\rm{s}}^2}\)
The muzzle velocity of the bullet is found by using the first equation of motion and by considering the initial velocity of bullet as zero.
The muzzle velocity of ball is given as
\(v = u + at\)
Here, \(v\) is the muzzle velocity of bullet.
Substitute all the values in the above equation.
\(\begin{array}{l}v = 0 + \left( {6.20 \times {{10}^5}\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {8.10 \times {{10}^{ - 4}}\;{\rm{s}}} \right)\\v = 502.2\;{\rm{m}}/{\rm{s}}\end{array}\)
Therefore, the muzzle velocity of bullet is \(502.2\;{\rm{m}}/{\rm{s}}\).