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 = (-4.35 times 10^7)t^2 + (3.15 times 10^5)t, where v is in meters 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 all numerical coefficients to exactly 3 significant figures.) a = m/s^2 x = m (b) Determine the length of time the bullet is accelerated. s (c) Find the speed at which the bullet leaves the barrel. m/s (d) What is the length of the barrel? m

Explanation / Answer

speed of bullet down the barrel opening is given by

v = (-4.35*10^7)t^2 + (3.15*10^5)t

v is in m/s , t is in s

acceleration of bullet when it leaves barrel is 0

a) v = (-4.35*10^7)t^2 + (3.15*10^5)t = dx/dt

integrating

x = -1.45*10^7*t^3 + 1.575*10^5t^2 + Co [ where Co is a constant ]

at t = 0, x = 0

so Co = 0

so, x = -1.45*10^7*t^3 + 1.575*10^5t^2

similiarly

a = dv/dt = -8.7*10^7t + 3.15*10^5

b) acceleration is 0 when bullet leavs the barrel

0 = -8.7*10^7t + 3.15*10^5

t = 0.00362 s

c) at t = 0.00362

v = 570.2586 m/s

d) length of barrel = a at t = 0.00362 s

x = 1.376093044 m

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