Can you please show your work? Thanks The speed of a bullet as it travels down t

Question

Can you please show your work?

Thanks

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (-4.05 x io7) t 2 + (2.40 x io5) t, where / is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. 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.) Determine the length of time the bullet is accelerated. Find the speed at which the bullet leaves the barrel, m/s What is the length of the barrel? m

Explanation / Answer

a)

acceleration, a = dv/dt = (-8.1* 10^7)*t + (2.4 * 10^5) m/s^2

position, x = integration of v with respect to t

x = (-1.35 * 10^7)*t^3 + (1.2* 10^5)*t^2 m

b)

a = 0 when bullet leaves barrel

so,

(-8.1* 10^7)*t + (2.4 * 10^5) = 0

so,

t = 0.002963 seconds = 0.003 seconds

c) leaving velocity = v(t=0.003)

v = (-4.05 * 10^7)*t^2 + (2.4 * 10^5)*t

v = 355.6 m/s

d) length of barrel = xpoition of bullet at time t = 0.002963 seconds

length = 0.7023 m = 0.702 m

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