The speed of a bullet as it travels down the barrel of a rifle toward the openin

Question

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v (-4.70 x 10') t (2.80 x 105) t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (Use t as necessary and round a numerical coefficients to exactly 3 significant figures.) m/s a (b) Determine the length of time the bullet is accelerated (c) Find the speed at which the bullet leaves the barrel. m/s (d) What is the length of the barrel?

Explanation / Answer

here ,

speed of bullet , v= -4.7 *10^7 t^2 + 2.80 *10^5 t

a)

as acceleration of bullet ,

a = dv/dt

a = d/dt(-4.7 *10^7 t^2 + 2.80 *10^5 t )

a = -9.4 *10^7 t + 2.80 *10^5 m/s^2

--------------------

x = integration (v * dt)

x = integration (-4.7 *10^7 t^2 + 2.80 *10^5 t )

x = - 1.567*10^7 t^3 + 1.4 *10^5 t^2

part b)

let the time is t

for a = 0

-9.4 *10^7 t + 2.80 *10^5 = 0

t = 2.98 *10^-3 s

the time for which bullet accelerate is 2.98 *10^-3 s

c)

Now, speed of bullet

v = -4.7 *10^7 t^2 + 2.80 *10^5 t

at t = 2.98 *10^-3 s

v = -4.7 *10^7 (2.98 *10^-3)^2 + 2.80 *10^5 (2.98 *10^-3)

v = 417.02 m/s

the speed at which the bullet leaves the barrel is 417.02 m/s

d)

length of barrel = 1.567*10^7 t^3 + 1.4 *10^5 t^2

length of barrel = 1.567*10^7 (2.98 *10^-3)^3 + 1.4 *10^5 (2.98 *10^-3)^2

length of barrel = 1.66 m

the length of barrel is 1.66 m

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