An antique rifle fires 9 mm bullets such that as they travel down the barrel of the rifle their speed is given by v = (−4.60 ✕ 107)t2 + (2.55 ✕ 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 (in m/s2) and position (in m) of the bullet as a function of time when the bullet is in the barrel. (Use the following as necessary: t. Round all numerical coefficients to at least three significant figures. Do not include units in your answers. Assume that the position of the bullet at t = 0 is zero.)

When The solution to this problem uses the concept of calculus and also motion
When The acceleration of the bullet is simply the result of differentiating than the velocity function concerning time.
Let's come up with equations for distance and acceleration as functions of time.
Now, From the definition of acceleration we know thata=dvdt then Taking the derivative of v concerning t yield sa=dvdt=(−5.15∗107)∗2∗t+(2.30∗105) From the definition of distance,
When we know that v=dxdt→dx=vdt Integrating velocity yield sx=(−5.15∗107)∗(t33)+(2.30∗105)∗(t22)+x0 where x0 is the starting position.
When If acceleration is zero when the bullet leaves the barrel, then we can use our equation for acceleration to determine the time the bullet is in the barrel.
Although when This is seen as0=(−5.15∗107)∗2∗t+(2.30∗105)→t=−(2.30∗105)(−5.15∗107) Knowing the time, also we can solve for the velocity as the bullet leaves the barrel by plugging time back into our given equation for velocity.
Then The length of the barrel can be solved by plugging time back into our equation for distance and also setting x0=0.

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