The-velocity-of-a-particle-moving-in-straight-line-is-given-by-the-graph-shown-here-Draw-the-acceleration-position-graph-

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 .

Commented by Tinkutara last updated on 23/May/17

Answered by mrW1 last updated on 23/May/17

a = \frac{d v}{d t} = \frac{d v}{d x} \times \frac{d x}{d t} = v \frac{d v}{d x}

v = v_{0} (1 - \frac{x}{x_{0}})

\frac{d v}{d x} = - \frac{v_{0}}{x_{0}}

\Rightarrow a = - \frac{v_{0}}{x_{0}} v = - \frac{v_{0}}{x_{0}} \times v_{0} (1 - \frac{x}{x_{0}}) = \frac{v_{0}^{2}}{x_{0}} (\frac{x}{x_{0}} - 1)

a t x = 0 : a = - \frac{v_{0}^{2}}{x_{0}}

a t x = x_{0} : a = 0

Commented by mrW1 last updated on 23/May/17

Commented by Tinkutara last updated on 23/May/17

Thanks!

Answered by ajfour last updated on 23/May/17

v \frac{d v}{d s} = a

\frac{d v}{d s} = \frac{a}{v} = - \frac{v_{0}}{x_{0}}

a = - \frac{v_{0}}{x_{0}} v = - \frac{v_{0}}{x_{0}} (v_{0}) (1 - \frac{x}{x_{0}})

a = - {(\frac{v_{0}}{x_{0}})}^{2} (x_{0} - x) .

Commented by Tinkutara last updated on 23/May/17

Thanks!

Commented by mrW1 last updated on 23/May/17

C a n y o u d r a w t h e p o s i t i o n - t i m e

c u r v e p l e a s e ?

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}})

Commented by mrW1 last updated on 23/May/17

t h a t m e a n s t h e p o i n t x = x_{0} c a n n e v e r

b e r e a c h e d, s i n c e x (t) < x_{0}

Commented by ajfour last updated on 23/May/17

y e s, i b e l i e v e t h e s a m e . .

Facebook Tweet Pin