A) \[{{t}^{3/4}}\]
B) \[{{t}^{3/2}}\]
C) \[{{t}^{1/4}}\]
D) \[{{t}^{1/2}}\]
Correct Answer: B
Solution :
| [b] In case of velocity changing with distance, we apply integration method to determine the relationship between distance and time, if force is given. |
| As, power, P = constant |
| \[\Rightarrow \] \[Fv=P\] (\[\because \] P = force \[\times \] velocity) |
| \[\Rightarrow \] \[Ma\times v=P\] \[(\because F=Ma)\] |
| \[\Rightarrow \] \[va=\frac{P}{M}\] |
| \[\Rightarrow \] \[v\times \left[ \frac{v\,dv}{ds} \right]=\frac{P}{M}\] \[\left( \because a=\frac{v\,dv}{ds} \right)\] |
| \[\Rightarrow \] \[\int_{0}^{v}{{{v}^{2}}}dv=\int_{0}^{s}{\frac{P}{M}}\,ds\] |
| (assuming at t = 0 it starts from rest, i.e., from s = 0) |
| \[\Rightarrow \] \[\frac{{{v}^{3}}}{s}=\frac{P}{M}s\] |
| \[\Rightarrow \] \[v={{\left( \frac{3P}{M} \right)}^{1/3}}{{s}^{1/3}}\] |
| \[\Rightarrow \] \[\frac{ds}{dt}=K\,{{s}^{1/3}}\] \[\left[ \because K={{\left( \frac{3P}{M} \right)}^{1/3}} \right]\] |
| \[\Rightarrow \] \[\int_{0}^{s}{\frac{ds}{{{s}^{1/3}}}=\int_{0}^{t}{K\,dt}}\] |
| \[\Rightarrow \] \[\frac{{{s}^{2/3}}}{2/3}=Kt\] |
| \[\Rightarrow \] \[{{s}^{2/3}}=\frac{2}{3}Kt\] |
| \[\Rightarrow \] \[s={{\left( \frac{2}{3}K \right)}^{3/2}}\times {{t}^{3/2}}\] |
| \[\Rightarrow \] \[s\propto {{t}^{3/2}}\] |
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