An object with a mass of $5 k g$ $5 kg$ is traveling at $6 \frac{m}{s}$ $6 m/s$ . If the object is accelerated by a force of $f (x) = 2 x^{2} - x$ $f(x) = 2x^2-x$ over $x \in [0, 9]$ $x in [0, 9]$ , where x is in meters, what is the impulse at $x = 3$ $x = 3$ ?

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An object with a mass of $5 k g$ $5 kg$ is traveling at $6 \frac{m}{s}$ $6 m/s$ . If the object is accelerated by a force of $f (x) = 2 x^{2} - x$ $f(x) = 2x^2-x$ over $x \in [0, 9]$ $x in [0, 9]$ , where x is in meters, what is the impulse at $x = 3$ $x = 3$ ?

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Answer 1 · 2016-02-04T21:25:18

let's write the impulse equation:
$F (x) = m d \frac{v}{d t}$ $F(x) = m dv/dt$ re-write in terms of x
$F (x) = m \frac{d^{2} x}{{d t}^{2}}$ $F(x) = m (d^2x)/dt^2$
we have a differential equation of the form:
$\frac{d^{2} x}{{d t}^{2}} = \frac{1}{m} F (x, \frac{d x}{d t})$ $(d^2x)/dt^2 = 1/mF(x, dx/dt)$
Let $u = \frac{d x}{d t};$ $u = dx/dt;$
using chain rule $\frac{d u}{d t} = \frac{d u}{d x} \frac{d x}{d t} = u \frac{d u}{d x}$ $(du)/dt = (du)/dx (dx)/dt = u (du)/dx$
$u \frac{d u}{d x} = \frac{1}{m} F (x, u) = \frac{1}{m} f (x)$ $u (du)/dx = 1/mF(x, u) = 1/m f(x)$
Note now the Force function is a function of x ONLY.
$u \frac{d u}{d x} = \frac{1}{m} f (x) = 2 x^{2} - x$ $u (du)/dx = 1/m f(x) = 2x^2 -x$ rearrange and solve by integrating
$\int u d u = \frac{1}{m} \int (2 x^{2} - x) d x$ $int udu = 1/m int(2x^2 - x) dx$
$\frac{1}{2} u^{2} = \frac{1}{m} (\frac{2}{3} x^{3} - \frac{1}{2} x^{2}) + C$ $1/2u^2 = 1/m (2/3x^3 - 1/2 x^2) + C$
The velocity after replacing u by v is:
$v = \sqrt{\frac{2}{5} (\frac{2}{3} x^{3} - \frac{1}{2} x^{2}) + C}$ $v = sqrt(2/5 (2/3x^3 - 1/2 x^2) + C$
evaluate C by setting v(0) = 6
$v (0) = 6 = \sqrt{C}; C = 36$ $v(0) = 6 = sqrt(C); C = 36$
now calculate v(3)

Answer 2 · 2016-02-04T21:56:18

let's write the impulse equation:
$F (x) = m d \frac{v}{d t}$ $F(x) = m dv/dt$ re-write in terms of x
$F (x) = m \frac{d^{2} x}{{d t}^{2}}$ $F(x) = m (d^2x)/dt^2$
we have a differential equation of the form:
$\frac{d^{2} x}{{d t}^{2}} = \frac{1}{m} F (x, \frac{d x}{d t})$ $(d^2x)/dt^2 = 1/mF(x, dx/dt)$
Let $u = \frac{d x}{d t};$ $u = dx/dt;$
using chain rule $\frac{d u}{d t} = \frac{d u}{d x} \frac{d x}{d t} = u \frac{d u}{d x}$ $(du)/dt = (du)/dx (dx)/dt = u (du)/dx$
$u \frac{d u}{d x} = \frac{1}{m} F (x, u) = \frac{1}{m} f (x)$ $u (du)/dx = 1/mF(x, u) = 1/m f(x)$

Note now the Force function is a function of x ONLY.
$u \frac{d u}{d x} = \frac{1}{m} f (x) = 2 x^{2} - x$ $u (du)/dx = 1/m f(x) = 2x^2 -x$ rearrange and solve by integrating
$\int u d u = \frac{1}{m} \int (2 x^{2} - x) d x$ $int udu = 1/m int(2x^2 - x) dx$
$\frac{1}{2} u^{2} = \frac{1}{m} (\frac{2}{3} x^{3} - \frac{1}{2} x^{2}) + C$ $1/2u^2 = 1/m (2/3x^3 - 1/2 x^2) + C$
since the object started with a velocity of 6m/s
$C = 36;$ $C = 36;$ and initial velocity, $v (0) = \sqrt{C} = 6$ $v(0) = sqrt(C) = 6$
$v = \sqrt{\frac{2}{5} (\frac{2}{3} x^{3} - \frac{1}{2} x^{2})} + 6$ $v = sqrt(2/5 (2/3x^3 - 1/2 x^2)) + 6$

now calculate v(3) = 8.323 m/s
And the Impulse, $I = mDelta v = 5 * 2.323 = 11.62 N$

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An object with a mass of $5 k g$ $5 kg$ is traveling at $6 \frac{m}{s}$ $6 m/s$ . If the object is accelerated by a force of $f (x) = 2 x^{2} - x$ $f(x) = 2x^2-x$ over $x \in [0, 9]$ $x in [0, 9]$ , where x is in meters, what is the impulse at $x = 3$ $x = 3$ ?

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An object with a mass of $5 k g$ $5 kg$ is traveling at $6 \frac{m}{s}$ $6 m/s$ . If the object is accelerated by a force of $f (x) = 2 x^{2} - x$ $f(x) = 2x^2-x$ over $x \in [0, 9]$ $x in [0, 9]$ , where x is in meters, what is the impulse at $x = 3$ $x = 3$ ?