13. 0/5 points I Previous Answers SerPSE9 2.P059 My Notes Ask Your The speed of

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13. 0/5 points I Previous Answers SerPSE9 2.P059 My Notes Ask Your The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v-(-5.20% 10') t 2 + (2.95 × 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 all numerical coefficients to exactly 3 significant figures.) m/s (b) Determine the length of time the bullet is accelerated 0.0295 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. s (c) Find the speed at which the bullet leaves the barrel 435.125 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (d) What is the length of the barrel? 0856 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m

Explanation / Answer

a) Acceleration is given by dv/dt

Given, v = (-5.20*10^7) t2 + (2.95*10^5) t

So, acceleration, a= dv/dt = (-10.4*10^7)t + (2.95*10^5) m/s/s

The bullet's displacement (position in the barrel) is just the integral of its velocity:

x = v·dt = (1/3)(-5.20×10^7)t^3 + (1/2)(2.95×10^5)t^2+ C

(To figure out the constant "C" in the last formula, set t = 0:

x(0) = (1/3)(-5.20×10^7)(0)^3 + (1/2)(2.95×10^5)(0)^2 + C

x(0) = C

That means "C" is the bullet's location at t=0, which we can take to be zero. So C=0. The acceleration of the bullet becomes zero just as it reaches the end of the barrel.

a(t0) = 0 = (-10.4*10^7)t + (2.95*10^5)

and:

x(t0) = L= (1/3)(-5.20×10^7)(t)^3 + (1/2)(2.95×10^5)(t)^2

b) a = 0 = (-10.4*10^7)t + (2.95 x 10^5)

t = 0.0028 sec

c) v = (-5.20x 10^7)t^2 + (2.95 x 10^5)t @ t= 0.0028 sec

= (-5.20*10^7)*0.0028^2 + (2.95*10^5)0.0028

v = 418.32 m/s

d) L = (1/3)(-5.20×10^7)(0.0028)^3 + (1/2) (2.95×10^5)(0.0028)^2 = 0.78 m

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