Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A particle leaves the origin with an initial velocity\[\overrightarrow{u}=(3\hat{i})m{{s}^{-1}}\]and at a constant acceleration\[\overrightarrow{a}=(-1.0\hat{i}-0.5\hat{j})m{{s}^{-2}}\]. Its velocity v and position vector\[\overrightarrow{r}\]when it reaches its maximum\[x-\]coordinate are (1) \[v=-2\hat{j}m{{s}^{-1}}\] (2) \[v=(-1.5\hat{j})m{{s}^{-1}}\] (3) \[\overrightarrow{r}=(4.5\hat{i}-1.25\hat{j})m\] (4) \[\overrightarrow{r}=(4.5\hat{i}-2.25\hat{j})m\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: C
Solution :
\[{{u}_{x}}=3m{{s}^{-1}},{{a}_{x}}=-1.0\,m{{s}^{-2}}\] \[\therefore \]\[v_{x}^{2}=u_{x}^{2}+2{{a}_{x}}x\]or\[0={{(3)}^{2}}+2(-1)(x)\] or \[x=4.5m\] Also \[{{v}_{x}}={{u}_{x}}+{{a}_{x}}t\] \[0=3-1.0t\]or \[t=3s\] \[y={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}=0+\frac{1}{2}(-0.5){{(3)}^{2}}-2.25\,m\] and \[{{v}_{y}}={{a}_{y}}t=(-0.5)(3)=-1.5\,m{{s}^{-1}}\] \[\therefore \]\[v={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j}=0-1.5\hat{j}=(-1.5\hat{j})m{{s}^{-1}}\] and \[\overrightarrow{r}=x\hat{i}+y\hat{j}=(4.5\hat{j}-2.25\hat{j})m\]
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