A) \[\frac{2}{17}\text{rad}\,{{\text{s}}^{-1}}\]
B) \[\frac{1}{14}\text{rad}\,{{\text{s}}^{-1}}\]
C) \[\frac{4}{7}\text{rad}\,{{\text{s}}^{-1}}\]
D) \[\frac{6}{5}\text{rad}\,{{\text{s}}^{-1}}\]
Correct Answer: A
Solution :
[a] : Given, \[\vec{r}=\{(2t)\overset{\wedge }{\mathop{i}}\,+(2{{t}^{2}})\overset{\wedge }{\mathop{j}}\,\}m\] Comparing it with standard equation of position vector, \[\vec{r}=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,,\]we get \[x=2t\] and \[y=2{{t}^{2}}\] \[\Rightarrow \]\[{{v}_{x}}=\frac{dx}{dt}=2\]and\[{{v}_{y}}=\frac{dy}{dt}=4t\] \[\therefore \]\[\tan \theta =\frac{{{v}_{y}}}{{{v}_{x}}}=\frac{4t}{2}=2t\] Differentiating with respect to time we get, \[(se{{c}^{2}}\theta )\frac{d\theta }{dt}=2\] or\[(1+ta{{n}^{2}}\theta )\frac{d\theta }{dt}=2\]or\[(1+4{{t}^{2}})\frac{d\theta }{dt}=2\] or\[\frac{d\theta }{dt}=\frac{2}{1+4{{t}^{2}}}\] at\[t=2s,\left( \frac{d\theta }{dt} \right)=\frac{2}{1+4{{(2)}^{2}}}=\frac{2}{17}\text{rad}\,{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
