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 -(-6.00 × 107)t + (3.00 × 10% 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 all numerical coefficients to exactly 3 significant figures.) xm/s2 (b) Determine the length of time the bullet is accelerated. Your response differs from the correct answer by more than 10%. Double check your calculations. S (c) Find the speed at which the bullet leaves the barrel. x m/s (d) What is the length of the barrel? physPad

Explanation / Answer

given velocity v = (-6*10^7)t^2 + (3*10^5)t

(a)

acceleration = rate of change in velocity

a = dv/dt

a = (-6*10^7*2*t) + (3*10^5)       <<<-----ANSWER

position x = integration v*dt

x = integration (-6*10^7*t^2) + (3*10^5*t) dt

x = (-6*10^7*t^3/3) + (3*10^5*t^2/2)

x = (-2.00*10^7*t^3) + (1.50*10^5*t^2)     <<<-----ANSWER

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(b)

at the openeing the acceleration = 0

a = 0

(-6*10^7*2*t) + (3*10^5) = 0

t = 0.0025 s    <<<-----ANSWER

(c)

speed st which the bullet leaves the barrel v = speed at t = 0.0025 s

v = (-6*10^7*0.0025^2) + (3*10^5*0.0025)

v = 375 m/s     <<<-----ANSWER

(d)

length of the barrel

L = x at 0.0025

L = (-2.00*10^7*0.0025^3) + (1.50*10^5*0.0025^2)

L = 0.625 m <<<-----ANSWER

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