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Give answer! - A particle moves from the point , at t = 0, with an initial velocity . It is acted upon by a constan - Kinematics - JEE Main

A particle moves from the point \left ( 2.0\widehat{i}+4.0\widehat{j} \right )m, at t = 0, with an initial velocity \left ( 5.0\widehat{i}+4.0\widehat{j} \right )ms^{-1}. It is acted upon by a constant force which produces a constant force which produces a constant acceleration \left ( 4.0\widehat{i}+4.0\widehat{j} \right )ms^{-2}. What is the distance of the particle from the origin at time 2s?

 

  • Option 1)

     

    5 m

  • Option 2)

     

    10\sqrt{2}m

  • Option 3)

     

    15 m

  • Option 4)

     

    20\sqrt{2}m

Answers (1)
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A admin

 

2nd equation or Position- time equation -

s= ut +\frac{1}{2}at^{2}

s\rightarrow Displacement

u\rightarrowInitial velocity

a\rightarrowacceleration

t\rightarrow time

 

-

 

S=ut+\frac{1}{2}at^{2}

t=25

r_{i}=2\hat{i}+4\hat{j}

s_{i}=\vec{r}_{f}-\vec{r_{i}}

u=s\hat{i}+4\hat{j}

a=4\hat{i}+4\hat{j}

s=(s\hat{i}+4\hat{j})2+\frac{1}{2}\times 4\times (4\hat{i}+4\hat{j})

=10\hat{i}+8\hat{j}+8\hat{i}+8\hat{j}

s=18\hat{i}+16\hat{j}

\vec{r_{f}}=20\hat{i}+20\hat{j}

|\vec{r_{f}}|=20\sqrt{2}

 

 


Option 1)

 

5 m

Option 2)

 

10\sqrt{2}m

Option 3)

 

15 m

Option 4)

 

20\sqrt{2}m

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