A) \[\frac{2as}{m}{{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}}\]
B) \[\frac{2as}{m}{{\left( 1-\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}}\]
C) \[\frac{2a{{s}^{2}}}{mR}\]
D) \[\frac{2as}{m}\]
Correct Answer: A
Solution :
According to given problem \[\frac{1}{2}m{{v}^{2}}=a{{s}^{2}}\] \[v=s\sqrt{\frac{2a}{m}}\] So \[{{a}_{R}}=\frac{{{v}^{2}}}{R}=\frac{2a{{s}^{2}}}{mR}\] Furthermore as \[{{a}_{t}}=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}\] \[{{a}_{t}}=\left[ s\sqrt{\frac{2a}{m}} \right]\left[ \sqrt{\frac{2a}{m}} \right]\] \[\left[ \because v=s\sqrt{\frac{2a}{m}}\text{ and }\frac{dv}{ds}=\sqrt{\frac{2a}{m}} \right]\] \[{{a}_{t}}=\frac{2as}{m}\] Acceleration \[a=\sqrt{a_{R}^{2}+a_{t}^{2}}\] \[=\sqrt{\left[ \frac{2a{{s}^{2}}}{mR} \right]+{{\left[ \frac{2as}{m} \right]}^{2}}}\] \[=\frac{2as}{m}\sqrt{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}\]
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