A) \[2\text{ }{{m}^{-1}}\text{ }{{s}^{2}}\]
B) \[4\text{ }{{m}^{-1}}\text{ }{{s}^{2}}\]
C) \[7.2\text{ }{{m}^{-1}}\text{ }{{s}^{2}}\]
D) \[72\text{ }{{m}^{-1}}\text{ }{{s}^{2}}\]
Correct Answer: C
Solution :
For motion with uniform acceleration\[{{a}_{1}}\]: From v = u + at \[v={{a}_{1}}{{t}_{1}}(\because u=0)\] \[\therefore {{t}_{1}}=\frac{v}{{{a}_{1}}}\] ?(i) And from \[s=ut+\frac{1}{2}a{{t}^{2}}\], \[{{s}_{1}}=\frac{1}{2}{{a}_{1}}{{\left( \frac{v}{{{a}_{1}}} \right)}^{2}}=\frac{{{v}^{2}}}{2{{a}_{1}}}\] ?(ii) For motion with uniform retardation \[{{a}_{2}}\]: From \[v=u+at\] \[v={{a}_{2}}{{t}_{2}}(\because v-0,u=v,a=-{{a}_{2}})\] \[\therefore {{t}_{2}}=\frac{v}{{{a}_{2}}}\] ?(iii) And from \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[{{s}_{2}}=v\frac{v}{{{a}_{2}}}+\frac{1}{2}(-{{a}_{2}})\cdot \frac{{{v}^{2}}}{a_{2}^{2}}\] \[{{s}_{2}}=\frac{{{v}^{2}}}{{{a}_{2}}}-\frac{{{v}^{2}}}{2{{a}_{2}}}=\frac{{{v}^{2}}}{2{{a}_{2}}}\] ?(iv) Given, \[{{s}_{1}}+{{s}_{2}}=4km\] and \[{{t}_{1}}+{{t}_{2}}=4\]min \[\therefore \frac{{{v}^{2}}}{2}\left( \frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}} \right)=4\] ?(v) And \[v\left( \frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}} \right)=4\] ?(vi) Dividing eqn. (v) by eqn. (vi), we get v=2 Putting this in eqn. (vi) \[\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}=2\frac{{{\min }^{2}}}{km}=\frac{2\times 3600}{1000}\frac{{{s}^{2}}}{m}=7.2{{m}^{-1}}{{s}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
