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Question Number 13745 by Tinkutara last updated on 23/May/17
The velocity of a particle moving in  straight line is given by the graph shown  here. Draw the acceleration position  graph.
$$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in} \\ $$$$\mathrm{straight}\:\mathrm{line}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{shown} \\ $$$$\mathrm{here}.\:\mathrm{Draw}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{position} \\ $$$$\mathrm{graph}. \\ $$
Commented by Tinkutara last updated on 23/May/17
Answered by mrW1 last updated on 23/May/17
a=(dv/dt)=(dv/dx)×(dx/dt)=v(dv/dx)  v=v_0 (1−(x/x_0 ))  (dv/dx)=−(v_0 /x_0 )  ⇒a=−(v_0 /x_0 )v=−(v_0 /x_0 )×v_0 (1−(x/x_0 ))=(v_0 ^2 /x_0 )((x/x_0 )−1)  at x=0: a=−(v_0 ^2 /x_0 )  at x=x_0 : a=0
$${a}=\frac{{dv}}{{dt}}=\frac{{dv}}{{dx}}×\frac{{dx}}{{dt}}={v}\frac{{dv}}{{dx}} \\ $$$${v}={v}_{\mathrm{0}} \left(\mathrm{1}−\frac{{x}}{{x}_{\mathrm{0}} }\right) \\ $$$$\frac{{dv}}{{dx}}=−\frac{{v}_{\mathrm{0}} }{{x}_{\mathrm{0}} } \\ $$$$\Rightarrow{a}=−\frac{{v}_{\mathrm{0}} }{{x}_{\mathrm{0}} }{v}=−\frac{{v}_{\mathrm{0}} }{{x}_{\mathrm{0}} }×{v}_{\mathrm{0}} \left(\mathrm{1}−\frac{{x}}{{x}_{\mathrm{0}} }\right)=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{x}_{\mathrm{0}} }\left(\frac{{x}}{{x}_{\mathrm{0}} }−\mathrm{1}\right) \\ $$$${at}\:{x}=\mathrm{0}:\:{a}=−\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{x}_{\mathrm{0}} } \\ $$$${at}\:{x}={x}_{\mathrm{0}} :\:{a}=\mathrm{0} \\ $$
Commented by mrW1 last updated on 23/May/17
Commented by Tinkutara last updated on 23/May/17
Thanks!
$$\mathrm{Thanks}! \\ $$
Answered by ajfour last updated on 23/May/17
v(dv/ds)=a  (dv/ds)=(a/v)=−(v_0 /x_0 )  a=−(v_0 /x_0 )v =−(v_0 /x_0 )(v_0 )(1−(x/x_0 ))    a=−((v_0 /x_0 ))^2 (x_0 −x) .
$${v}\frac{{dv}}{{ds}}={a} \\ $$$$\frac{{dv}}{{ds}}=\frac{{a}}{{v}}=−\frac{\boldsymbol{{v}}_{\mathrm{0}} }{\boldsymbol{{x}}_{\mathrm{0}} } \\ $$$$\boldsymbol{{a}}=−\frac{\boldsymbol{{v}}_{\mathrm{0}} }{\boldsymbol{{x}}_{\mathrm{0}} }\boldsymbol{{v}}\:=−\frac{\boldsymbol{{v}}_{\mathrm{0}} }{\boldsymbol{{x}}_{\mathrm{0}} }\left(\boldsymbol{{v}}_{\mathrm{0}} \right)\left(\mathrm{1}−\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}_{\mathrm{0}} }\right) \\ $$$$\:\:\boldsymbol{{a}}=−\left(\frac{\boldsymbol{{v}}_{\mathrm{0}} }{\boldsymbol{{x}}_{\mathrm{0}} }\right)^{\mathrm{2}} \left(\boldsymbol{{x}}_{\mathrm{0}} −\boldsymbol{{x}}\right)\:. \\ $$
Commented by Tinkutara last updated on 23/May/17
Thanks!
$$\mathrm{Thanks}! \\ $$
Commented by mrW1 last updated on 23/May/17
Can you draw the position−time  curve please?
$${Can}\:{you}\:{draw}\:{the}\:{position}−{time} \\ $$$${curve}\:{please}? \\ $$
Commented by ajfour last updated on 23/May/17
Commented by ajfour last updated on 23/May/17
x=x_0 (1−e^(−v_0 t/x_0 ) )
$$\boldsymbol{{x}}=\boldsymbol{{x}}_{\mathrm{0}} \left(\mathrm{1}−\boldsymbol{{e}}^{−\boldsymbol{{v}}_{\mathrm{0}} \boldsymbol{{t}}/\boldsymbol{{x}}_{\mathrm{0}} } \right) \\ $$
Commented by mrW1 last updated on 23/May/17
that means the point x=x_0  can never  be reached, since x(t)<x_0
$${that}\:{means}\:{the}\:{point}\:{x}={x}_{\mathrm{0}} \:{can}\:{never} \\ $$$${be}\:{reached},\:{since}\:{x}\left({t}\right)<{x}_{\mathrm{0}} \\ $$
Commented by ajfour last updated on 23/May/17
yes, i believe the same..
$${yes},\:{i}\:{believe}\:{the}\:{same}.. \\ $$

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