NTA Abhyas 2020

NTA Abhyas 2020 · Physics · Motion in One Dimension

A particle moves with constant acceleration along a straight line....

A particle moves with constant acceleration along a straight line. If

{v}_{1}

{v}_{2}

and

{v}_{3}

are the average velocities in the three successive intervals

{t}_{1}

{t}_{2}

and

{t}_{3}

of time, then the correct relation is

Options

A $\dfrac{{v}_{1} - {v}_{2}}{{v}_{2} - {v}_{3}} = \dfrac{{t}_{1} - {t}_{2}}{{t}_{2} + {t}_{3}}$
B $\dfrac{{v}_{1} - {v}_{2}}{{v}_{2} - {v}_{3}} = \dfrac{{t}_{1} - {t}_{2}}{{t}_{1} - {t}_{3}}$
C $\dfrac{{v}_{1} - {v}_{2}}{{v}_{2} - {v}_{3}} = \dfrac{{t}_{1} - {t}_{2}}{{t}_{2} - {t}_{3}}$
D $\dfrac{{v}_{1} - {v}_{2}}{{v}_{2} - {v}_{3}} = \dfrac{{t}_{1} + {t}_{2}}{{t}_{2} + {t}_{3}}$

D. $\dfrac{{v}_{1} - {v}_{2}}{{v}_{2} - {v}_{3}} = \dfrac{{t}_{1} + {t}_{2}}{{t}_{2} + {t}_{3}}$

Let

u

be the initial velocity. (\therefore v_{1}=u+a t_{1}, v_{2}=u+d

\left(

t_{3}+t_{2}\right)

\mathrm{and}

) (v_{3}=u+a

\left(

t_{1}+t_{3}+t_{3}\right)) Now, (v_{1}=

\frac{u w_{1}}{2}, \frac{u\left(u+a t_{1}\right)}{2}=

\frac{1}{2}

) at, (v_{1}=

\frac{v_{1}+c_{2}}{2}+

u+a t_{1}+

\frac{1}{2}

a t_{2}) (v_{3}=

\frac{v_{2}+v_{2}}{2}=

u+a t_{1}+a t_{2}+

\frac{1}{2}

a t_{3}) So, (

\left(

y_{1}-v_{2}\right)=-

\frac{1}{2}

a\left(

t_{3}+t_{2}\right)) and (

\left(

v_{2}-v_{3}\right)=-

\frac{1}{2}

) a (\left(

t_{2}+t_{3}\right)) (\therefore

\frac{\left(v_{1}-v_{2}\right)}{\left(v_{2}-v_{3}\right)}=\frac{\left(t_{1}+t_{2}\right)}{\left(t_{2}+t_{3}\right)}

)