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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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