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A body of mass $0.40 \mathrm{~kg}$ moving initially with a constant speed of $10 \mathrm{~m} / \mathrm{s}^{-1}$ to the north is subject to a constant force of $8.0 \mathrm{~N}$ directed towards the south for $30 \mathrm{~s}$. Take the instant the force is applied to be $t=0$, and the position of the body at that time to be $x=0$, predict its position at $t=-5 \mathrm{~s}, 25 \mathrm{~s}, 100 \mathrm{~s}$ ?
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Here, $m=0.4 \mathrm{~kg}, u=10 \mathrm{~ms}^{-1}$ due north
$F=-8 \mathrm{~N}$, (negative sign shows the force directed opposite).
Therefore, $a=\frac{F}{m}=\frac{-8}{0.4}=-20 \mathrm{~ms}^{-2}(0 \leq \mathrm{t} \leq 30 \mathrm{~s})$
When $t=-5 s, x=u t=10(-5)=-50 \mathrm{~m}($ as $a=0)$
When $t=25 \mathrm{~s}, x=u t+1 / 2 a t^2=10 \times 25$
$+1 / 2(-20)(25)^2=-6000 \mathrm{~m}$
Upto $t=30 \mathrm{~s}$, motion is under acceleration, i.e.
$x=u t+1 / 2 a t^2=10 \times 30+1 / 2(-20)(30)^2=-8700 \mathrm{~m}$
At $t=30 \mathrm{~s}, v=u+a t=10-20 \times 30=-590 \mathrm{~m} / \mathrm{s}$
During $t=30$ to $100 \mathrm{~s}, \quad x_2=v t=-590 \times 70$
$=-41300 \mathrm{~m} \quad$ [as the force is removed $a=0$ ]
Total distance, $x_1+x_2=-(8700+41300) \mathrm{m}$
$=-50 \mathrm{~km}$
$F=-8 \mathrm{~N}$, (negative sign shows the force directed opposite).
Therefore, $a=\frac{F}{m}=\frac{-8}{0.4}=-20 \mathrm{~ms}^{-2}(0 \leq \mathrm{t} \leq 30 \mathrm{~s})$
When $t=-5 s, x=u t=10(-5)=-50 \mathrm{~m}($ as $a=0)$
When $t=25 \mathrm{~s}, x=u t+1 / 2 a t^2=10 \times 25$
$+1 / 2(-20)(25)^2=-6000 \mathrm{~m}$
Upto $t=30 \mathrm{~s}$, motion is under acceleration, i.e.
$x=u t+1 / 2 a t^2=10 \times 30+1 / 2(-20)(30)^2=-8700 \mathrm{~m}$
At $t=30 \mathrm{~s}, v=u+a t=10-20 \times 30=-590 \mathrm{~m} / \mathrm{s}$
During $t=30$ to $100 \mathrm{~s}, \quad x_2=v t=-590 \times 70$
$=-41300 \mathrm{~m} \quad$ [as the force is removed $a=0$ ]
Total distance, $x_1+x_2=-(8700+41300) \mathrm{m}$
$=-50 \mathrm{~km}$
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