Answer:
a = - 1.987 × 10⁶ ft/s²
t = 6.84 × 10⁻⁴ s
Explanation:
v₀ = 910 ft/s
x = 5 in.
relation v = v₀ - k x
v = 0 as body comes to rest
0 = 900 - 5k/12
k = 2184 s⁻¹
acceleration
[tex]\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
where
(A) a = -k × v
at v= 910 ft/s
a = - 1.987 × 10⁶ ft/s²
(B) at x = 3.9 in.
v = 910 - 3.9(2184)/12
v = 200.2 m/s
[tex]\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\frac{dv}{v} = -kdt[/tex]
[tex]\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt[/tex]
[tex]ln(200.2)-ln(900) = -kt[/tex]
t = 6.84 × 10⁻⁴ s
