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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?

Views

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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