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 = (-5.70 times 10^7)t^2 + (3.50 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. .00307 s Find the speed at which the bullet leaves the barrel. .05372 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully, m/s What is the length of the barrel? 1.649 Your response differs from the correct answer by more than 10%. Double check your calculations, m

Explanation / Answer

(a)

v = (-5.70 x 107)t2 + (3.50 x 105)t

a = dv/dt = (-11.4 x 107)t + (3.5 x 105) m/s^2

Integrate v to get x

x = (-5.70 x 107) t^3/3 + (3.50 x 105)t^2/2 + x0

(b)

(-11.4 x 107)t + (3.5 x 105) = 0

t = 0.00307 s

(c)

v = (-5.70 x 107)t2 + (3.50 x 105)t = (-5.70 x 107)(0.00307)2 + (3.50 x 105)(0.00307)

v = 537.34 m/s

(d)

x = (-5.70 x 107) (0.00307)^3/3 + (3.50 x 105)(0.00307)^2/2

=1.099 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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