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  • A particle starts from the origin at t= 0 s with a velocity of 10.0 hat j m/s and moves in the x-y plane with a constant acceleration of (8.0 hat i +2.0 hat j) m s^ -2. (a) At what time is the x coordinate of the particle 16 m?

Q. 4.21 A particle starts from the origin at t=0\; s  with a velocity of  10.0\; \hat{j}\; m/s and moves in the x-y plane with a constant acceleration of  \left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}.

(a) At what time is the x- coordinate of the particle 16 \; m? What is the y-coordinate of the particle at that time

Answers (1)

best_answer

We are given the velocity of the particle as  10.0\; \hat{j}\; m/s.

And the acceleration is given as :

                                                           \left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}

So, the velocity due to acceleration will be :

                                                              a\ =\ \frac{dv}{dt}

So,                                                         dv\ =\ \left ( 8.0\ \widehat{i}\ +\ 2.0\ \widehat{j} \right )dt

By integrating both sides,

or                                                           v\ =\ 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u

Here u is the initial velocity (at t = 0 sec).

Now,

                                                              v\ =\ \frac{dr}{dt}

or                                                        dr\ =\ \left ( 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u \right )dt

Integrating both sides,  we get

                                                            r\ =\ 8.0\times \frac{1}{2}t^2\ \widehat{i}\ +\ 2.0\times\frac{1}{2} t^2\ \widehat{j}\ +\ (10t\ ) \widehat{j}

or                                                         r\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}                              

or                                        x\widehat{i}\ +\ y\widehat{j}\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}

 

Comparing coefficients, we get  :

                                             x\ =\ 4t^2         and            y\ =\ 10t\ +\ t^2

In the question, we are given x = 16.

So                                        t   =   2 sec

and                                     y  =  10 (2) + 2 2  =   24 m.

 

Posted by

Devendra Khairwa

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