Answer:
Let
the point starts moving from O with a uniform acceleration a along a st. line.
It reaches at locations A, B and C at timings and respectively Fig. 2(HT).6.
Let
\[\upsilon '\],\[\upsilon ''\], \[\upsilon '''\]be the velocity of point at A, B and
C respectively
\[\therefore \] Initial velocity of point at O, u = 0
Using
the formula; \[\upsilon =u+at\]
Taking
motion from O to A, we have ;
\[\upsilon
'=0+a{{t}_{1}}=a{{t}_{1}}\]
Taking
motion from O to B, we have;
\[\upsilon
''=0+a({{t}_{1}}+{{t}_{2}})=a({{t}_{1}}+{{t}_{2}})\]
Taking
motion form O to C, we have
\[\upsilon
'''=0+a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})=a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})\]
\[\therefore \] Average velocity in interval of time \[{{t}_{1}}\].
\[{{\upsilon }_{1}}=\frac{0+\upsilon
'}{2}=\frac{a{{t}_{1}}}{2}\] ??
(1)
Average
velocity in interval of time \[{{t}_{2}}\],
\[{{\upsilon }_{2}}=\frac{\upsilon '+\upsilon
''}{2}=\frac{a{{t}_{1}}+a\left( {{t}_{1}}+{{t}_{2}}
\right)}{2}=a{{t}_{1}}+\frac{1}{2}a{{t}_{2}}\] ??.. (2)
Average
velocity in interval of time \[{{t}_{3}}\],
\[{{\upsilon }_{3}}=\frac{\upsilon ''+\upsilon
'''}{2}=\frac{a\left( {{t}_{1}}+{{t}_{2}} \right)+a\left(
{{t}_{1}}+{{t}_{2}}+{{t}_{3}} \right)}{2}=a\left( {{t}_{1}}+{{t}_{2}}
\right)+\frac{1}{2}a{{t}_{3}}\] ??. (3)
\[\therefore \] \[{{\upsilon }_{2}}-{{\upsilon
}_{1}}=\left( a{{t}_{1}}+\frac{1}{2}a{{t}_{2}}
\right)-\frac{a{{t}_{1}}}{2}=\frac{a}{2}\left( {{t}_{1}}+{{t}_{2}} \right)\] ??..
(4)
and
\[{{\upsilon }_{3}}-{{\upsilon }_{2}}=\left(
a({{t}_{1}}+{{t}_{2}})+\frac{1}{2}a{{t}_{3}} \right)-\left[
a{{t}_{1}}+\frac{1}{2}a{{t}_{2}} \right]=\frac{a}{2}\left( {{t}_{2}}+{{t}_{3}}
\right)\] ??. (5)
From
(4) and (5), \[\frac{{{\upsilon }_{2}}-{{\upsilon }_{1}}}{{{\upsilon
}_{3}}-{{\upsilon }_{2}}}=\frac{\left( {{t}_{1}}+{{t}_{2}} \right)}{\left(
{{t}_{2}}+{{t}_{3}} \right)}\]or \[\frac{{{\upsilon }_{1}}-{{\upsilon
}_{2}}}{{{\upsilon }_{2}}-{{\upsilon
}_{3}}}=\frac{{{t}_{1}}+{{t}_{2}}}{{{t}_{2}}+{{t}_{3}}}\]
You need to login to perform this action.
You will be redirected in
3 sec