Answer
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Hint: Calculate average velocity, we add initial and final velocity and divide their sum with 2, then to find acceleration we use the third equation of motion, and finally to find out the time taken by a bullet to travel, we use the first equation of motion.
Given:
Length of barrel, $S = 60cm$
As the velocity of bullet is given in $m/s$then, the distance should be in $m$
So, $S = 0.6m$
Final velocity, $V = 750m/s$
Initial velocity, $U = 0$
Formula used:
${V_{avg}} = \dfrac{{U + V}}{2}$
${V^2} = {U^2} + 2aS$
$\dfrac{{V - U}}{T} = a$
Complete Step by step solution:
To calculate the average velocity of the bullet, we use a formula in which the sum of initial and final velocity is divided by 2.
Now, we will calculate the average velocity
${V_{avg}} = \dfrac{{U + V}}{2}$
\[ = \dfrac{{0 + 750}}{2}\](Putting value of $U$and$V$)
$ = 375m/s$
Here we got an average velocity of the bullet which is $375m/s$
Now, you know there is a relationship between initial velocity, final velocity, distance, and acceleration
${V^2} = {U^2} + 2aS$(Here $a$is an acceleration)
${750^2} = 0 + 2a \times 0.6$(Putting values of $U,S$and$V$ )
$a = \dfrac{{562500}}{{1.2}} = 468750 = 46.88 \times {10^4}m/s$
Here we got acceleration, which we can use in the relation of final velocity, initial velocity, acceleration, and time, where we have everything except time, so we will calculate time from this relation.$\dfrac{{V - U}}{T} = a$
After simplification, we can write this equation, as below
$T = \dfrac{{V - U}}{a}$
$ = \dfrac{{750 - 0}}{{46.88 \times {{10}^4}}}$(Putting values of $U,a$and$V$)
$ = 15.998 \times {10^{ - 4}}s$
Here, we have calculated the time taken by a bullet to travel.
Note: Point to be noted is, as we know that at the starting point the bullet was in the rifle at rest position, so we will assume that the initial velocity of the bullet is zero. And we should note equations of motion to relate our given value to find out the acceleration of the bullet and time taken by the bullet.
Given:
Length of barrel, $S = 60cm$
As the velocity of bullet is given in $m/s$then, the distance should be in $m$
So, $S = 0.6m$
Final velocity, $V = 750m/s$
Initial velocity, $U = 0$
Formula used:
${V_{avg}} = \dfrac{{U + V}}{2}$
${V^2} = {U^2} + 2aS$
$\dfrac{{V - U}}{T} = a$
Complete Step by step solution:
To calculate the average velocity of the bullet, we use a formula in which the sum of initial and final velocity is divided by 2.
Now, we will calculate the average velocity
${V_{avg}} = \dfrac{{U + V}}{2}$
\[ = \dfrac{{0 + 750}}{2}\](Putting value of $U$and$V$)
$ = 375m/s$
Here we got an average velocity of the bullet which is $375m/s$
Now, you know there is a relationship between initial velocity, final velocity, distance, and acceleration
${V^2} = {U^2} + 2aS$(Here $a$is an acceleration)
${750^2} = 0 + 2a \times 0.6$(Putting values of $U,S$and$V$ )
$a = \dfrac{{562500}}{{1.2}} = 468750 = 46.88 \times {10^4}m/s$
Here we got acceleration, which we can use in the relation of final velocity, initial velocity, acceleration, and time, where we have everything except time, so we will calculate time from this relation.$\dfrac{{V - U}}{T} = a$
After simplification, we can write this equation, as below
$T = \dfrac{{V - U}}{a}$
$ = \dfrac{{750 - 0}}{{46.88 \times {{10}^4}}}$(Putting values of $U,a$and$V$)
$ = 15.998 \times {10^{ - 4}}s$
Here, we have calculated the time taken by a bullet to travel.
Note: Point to be noted is, as we know that at the starting point the bullet was in the rifle at rest position, so we will assume that the initial velocity of the bullet is zero. And we should note equations of motion to relate our given value to find out the acceleration of the bullet and time taken by the bullet.
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