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A particle starts from the origin at t equal to 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. (b) What is the speed of the particle at the time?

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}. 

(b) What is the speed of the particle at the time?

Answers (1)
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The velocity of particle is given by :

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

Put  t = 2 sec,

 So velocity becomes : 

                                                              v\ =\ 8.0(2)\ \widehat{i}\ +\ 2.0(2)\ \widehat{j}\ +\ 10\widehat{j}

or                                                           v\ =\ 16\ \widehat{i}\ +\ 14\ \widehat{j}

Now,  the magnitude of velocity gives :

                                                               \left | v \right |\ =\ \sqrt{16^2\ +\ 14^2}

                                                                       =\ \sqrt{256+196}

                                                                       =\ 21.26\ m/s

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