The speed of a bullet as it travels down the barrel of a rifletoward the opening

Question

The speed of a bullet as it travels down the barrel of a rifletoward the opening is given by v = (-4.65 x107) t 2 +(2.30 x 105) t, where vis in meters per second and t is in seconds. Theacceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position ofthe bullet as a function of time when the bullet is in the barrel.(Use t as necessary and round all numerical coefficientsto exactly 3 significant figures.) a = --------------------------m/s^2 x = ---------------------------m (b) Determine the length of time the bullet isaccelerated.
------------------s (c) Find the speed at which the bullet leavesthe barrel.
------------------------ m/s (d) What is the length of the barrel? ----------------------- m (a) Determine the acceleration and position ofthe bullet as a function of time when the bullet is in the barrel.(Use t as necessary and round all numerical coefficientsto exactly 3 significant figures.) a = --------------------------m/s^2 x = ---------------------------m (b) Determine the length of time the bullet isaccelerated.
------------------s (c) Find the speed at which the bullet leavesthe barrel.
------------------------ m/s (d) What is the length of the barrel? ----------------------- m

Explanation / Answer

Given velocity    v =(-4.65 x107) t 2 +(2.30 x 105) t Accleration a = dv / dt                      =  2(-4.65 x107) t+ (2.30 x 105)                    =  (-9.9 x107) t + (2.30 x 105) Displacement or position    x =integral v dt                                             =  [(-4.65 x107) t 3/ 3 ] + (2.30 x 105) [ t 2 / 2]                                             = [(-1.55 x107) t 3 ] + (1.15 x 105) [ t 2 ] Accleration = 0 at    end of thebarrel          (-9.9x107) t +(2.30 x 105) = 0                           t = 2.32 * 10 ^ -3 s Position at time t = 2.32 * 10 ^ -3 s  is    x =  [(-1.55 x107) (  2.32 * 10 ^ -3)  3 ] + (1.15 x 105) ( 2.32 * 10 ^-3) 2 ]          = - 0.19436 + 0.6199      = 0.4246 m So. length of the barallel L = 0.4246 m Accleration = 0 at    end of thebarrel          (-9.9x107) t +(2.30 x 105) = 0                           t = 2.32 * 10 ^ -3 s Position at time t = 2.32 * 10 ^ -3 s  is    x =  [(-1.55 x107) (  2.32 * 10 ^ -3)  3 ] + (1.15 x 105) ( 2.32 * 10 ^-3) 2 ]          = - 0.19436 + 0.6199      = 0.4246 m So. length of the barallel L = 0.4246 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.