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  • At time a particle starts travelling from a height <img alt="7 \hat{z} \mathrm{~cm}" src="https

At time \mathrm{t=0} a particle starts travelling from a height 7 \hat{z} \mathrm{~cm} in a plane keeping \mathrm{z} coordinate constant. At any instant of time it's position along the \mathrm{}\hat{x}$ and $\hat{y} directions are defined as 3 \mathrm{t}$ and $5 \mathrm{t}^{3} respectively. At \mathrm{t=1} s acceleration of the particle will be
 

Option: 1

\mathrm{-30 \: \hat{y}}


Option: 2

\mathrm{30 \hat{y}}
 


Option: 3

\mathrm{3 \hat{x}+15 \hat{y}}
 


Option: 4

\mathrm{3 \hat{x}+15 \hat{y}+7 \hat{z}}


Answers (1)

best_answer

\mathrm{\bar{r}=3 t \hat{i}+5 t^3 \hat{j}+7 \hat{k}}

\mathrm{Initial \: \: position \: \: is \: \: (0,0,7) \mathrm{cm}}

\mathrm{\bar{r} \rightarrow position \: vector}

\mathrm{\bar{v} =\frac{d \bar{r}}{d t}=3 \hat{i}+15 t^2 \hat{\jmath}+0 }

\mathrm{\bar{a} =\frac{d \bar{v}}{d t}=0+30 t \hat{\jmath}+0 }

\mathrm{at\: t=1,\bar{a}=30\hat{j}\: \: \: or}

                  \mathrm{\bar{a}=30\hat{y}}

Hence 2 is correct option







The comect option is (2)

Posted by

Gaurav

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