Question

a particle is uniformly accelerated from A to B, a distance of 192m, and is then uniformly retarded from B to C, a distance of 60m. the speeds of the particle at A and B are 4m/s and Vm/s respectively and the particle comes to rest at C. express, in terms of V only, the times taken by the particle to move from A to B and from A to C. Given that the total time taken by the particle to move from A to C is 22 seconds, find: (a) the value of V, (b) the acceleration and the retardation of the particle.

Answer 1

\(First\ you\ find\ the\ average\ speed\ of\ particle\ on\ each\ distance\ from\ A\ to\ B\\ and\ from\ B\ to\ C\)\\
\({ v }_{ A\rightarrow B }=\frac { { v }_{ A }+{ v }_{ B } }{ 2 } =\frac { 4+V }{ 2 } \\ { v }_{ B\rightarrow C }=\frac { { v }_{ B }{ +v }_{ C } }{ 2 } =\frac { V+0 }{ 2 }\)

\(Then\ find\ the\ time\ taken\ by\ the\ particle\ to\ move\ from\ A\ to\ B\ and\ from\ B\ to\ C\)
\({ t }_{ A\rightarrow B }=\frac { { s }_{ A\rightarrow B } }{ { v }_{ A\rightarrow B } } =\frac { 192 }{ \frac { 4+V }{ 2 } } =\frac { 384 }{ 4+V } \\ { t }_{ B\rightarrow C }=\frac { { s }_{ B\rightarrow C } }{ { v }_{ B\rightarrow C } } =\frac { 60 }{ \frac { V }{ 2 } } =\frac { 120 }{ V }\)

\(The\ total\ time\ taken\ by\ the\ particle\ is\ 22\ seconds,\ so:\)
\(t={ t }_{ A\rightarrow B }+{ t }_{ B\rightarrow C }=20\\ \Leftrightarrow \frac { 384 }{ 4+V } +\frac { 120 }{ V } =20\\ \Leftrightarrow 20{ V }^{ 2 }-424V-420=0\\ \Leftrightarrow V\approx 20\quad (m/s)\)