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 = (-5.65 times 10^7) t^2 + (2.30 times 10^5) 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. 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 Determine the length of time the bullet is accelerated. s Find the speed at which the bullet leaves the barrel. m/s What is the length of the barrel? m

Explanation / Answer

A.

your value of x is almost correct, as you forgot to add the x0 in the above expression.

As after integration there will be a constant.

so correct answer will be

x = -(5.65/3)*10^7*t^3 + (2.3/2)*10^5*t^2 + x0

B.

it means the interval when acceleration reached zero.

a = 0 = - 11.3*10^7*t + 2.3*10^5

t = 2.3*10^5/(11.3*10^7)

t = 2.03*10^-3 sec

so the bullet is accelerated for 2.03 mili sec

C.

velocity after 2.03*10^-3 sec

v = -5.65*10^7*t^2 + 2.3*10^5*t

at t = 2.03*10^-3 sec

v = -5.65*10^7*(2.03*10^-3)^2 + 2.3*10^5*2.03*10^-3

v = 234.06 m/sec

D.

displacement after 2.03 milisec

x = -(5.65/3)*10^7*t^3 + (2.3/2)*10^5*t^2 + x0

x = (x at 2.03 mili sec) - (x at t = 0 sec)

x = -5.65*10^7*(2.03*10^-3)^3/3 + 2.3*10^5*(2.03*10^-3)^2/2 + x0 - 0 - 0 - x0

x = 0.316 m

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