I need help with this problem guys please help me out :( The speed of a bullet a

Question

I need help with this problem guys please help me out :(

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (-4. 00 times 10 7) t 2 + (2. 20 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. ) Determine the length of time the bullet is accelerated. Find the speed at which the bullet leaves the barrel. What is the length of the barrel?

Explanation / Answer

a)a=dv/dt=-8*10^7t+2.2*10^5 @ t=3sec..a=-797.8*10^5 x=integralv =-4*1067*t^3/3+1.1*10^5*t^2 put t=3 b)-8*10^7t+2.2*10^5=0 we get t=2.75msec c)v @t=2.75msec=0 d)x @t=2.75msec put t in x eqn..

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