Question

A muzzle-loading rifle fires 22LR bullets such that as they travel down the barrel of the rifle their speed is given by v (-4.75 10t2(2.55 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 (in m/s) and position (in m) of the bullet as a function of time when the bullet is in the barrel. (Use the following as necessary: t Round all numerical coefficients to at least three significant figures. Do not include units in your answers. Assume that the position of the bullet at t- 0 is zero.) a(t) x(t) m/s2 (b) Determine the length of time the bullet is accelerated (in s). (c) Find the speed at which the bullet leaves the barrel (in m/s). m/s (d) What is the length of the barrel (in m)?

Answer 1

Given,

v = (-4.75 x 10^7)t^2 + (2.55 x 10^5)t

a)we know that

a = dv/dt and v = dx/dt => dx = v dt

we will diffrentiate the given equation of v for finding a and integrate for finding x

a = d( (-4.75 x 10^7)t^2 + (2.55 x 10^5)t)/dt

a = (-9.5 x 10^7) t + (2.55 x 10^5)

x = int ( (-4.75 x 10^7)t^2 + (2.55 x 10^5)t) dt

x = (-1.5833 x 10^7) t^3 + (1.275 x 10^5) t^2

b)for this, a = 0

(-9.5 x 10^7) t + (2.55 x 10^5) = 0

t = 2.55 x 10^5/(9.5 x 10^7) = 2.68 x 10^-3 s

Hence, t = 2.68 x 10^-3 s

c)put t = 2.68 x 10^-3 in eqn of v

v = (-4.75 x 10^7)(2.68 x 10^-3)^2 + (2.55 x 10^5)(2.68 x 10^-3) = 342.24 m/s

v = 342.24 m/s

d)we have

x = (-1.5833 x 10^7) t^3 + (1.275 x 10^5) t^2

put t = 2.68 x 10^-3

x = (-1.5833 x 10^7) (2.68 x 10^-3)^3 + (1.275 x 10^5) (2.68 x 10^-3)^2 = 0.61 m

Hence, x = 0.61 m