Question

A particle starts from the origin at t = 0 s with a velocity of $10.0\mathop j\limits^ \wedge m/s$ and moves in the x-y plane with a constant acceleration of $(8.0\mathop i\limits^ \wedge + 2.0\mathop j\limits^ \wedge )m/{s^2}$.
(a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
(b) What is the speed of the particle at the time?

Answer 1

Hint: It is given that the particle is starting from the rest and moving with constant acceleration along x and y axes. We apply kinematic equations in order to find out the time taken to reach particular coordinate and speed of the particle after the given time.

Formula used:
$\eqalign{
  & s = ut + \dfrac{1}{2}a{t^2} \cr
  & v = u + at \cr} $

Complete step-by-step answer:
Velocity is the rate of change of displacement with time and acceleration is rate of change of velocity with time. The kinematic equations we have are valid only for the one directional motion so we calculate every value along x and y directions separately.
If we see clearly along the x axis there is no initial velocity and we have to find the time taken for the particle to reach x coordinate of 16m.
The formula is $s = ut + \dfrac{1}{2}a{t^2}$
Where ‘s’ is the displacement and ‘u’ is the initial velocity and ‘a’ is the acceleration
Acceleration is given as $(8.0\mathop i\limits^ \wedge + 2.0\mathop j\limits^ \wedge )m/{s^2}$ i.e along x axis its component Is 8. Along the x axis the initial velocity is zero. Displacement travelled along the x axis is 16m. by substituting all these values in $s = ut + \dfrac{1}{2}a{t^2}$ we get
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow 16 = \dfrac{1}{2} \times 8 \times {t^2} \cr
  & \Rightarrow 16 = 4 \times {t^2} \cr
  & \Rightarrow 2 = t \cr
  & \Rightarrow t = 2 \cr} $
So with in two seconds particle reach x coordinate 16m
Now y coordinate at 2 sec will be $s = ut + \dfrac{1}{2}a{t^2}$ initial y component velocity is given as $10.0\mathop j\limits^ \wedge m/s$ and the acceleration component along y direction is 2 which is obtained from acceleration value.
So we have
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow s = 10 \times 2 + \dfrac{1}{2} \times 2 \times {2^2} \cr
  & \Rightarrow s = 20 + 4 \cr
  & \Rightarrow s = 24m \cr} $
So y coordinate will be 24 meter.
To find out the velocity at 2 second we have the formula $v = u + at$
Along x axis velocity will be
$v = u + at$
$\eqalign{
  & \Rightarrow v = 0 + 8 \times 2 \cr
  & \Rightarrow v = 16m/s \cr} $
Along y axis velocity will be
$v = u + at$
$\eqalign{
  & \Rightarrow v = 10 + 2 \times 2 \cr
  & \Rightarrow v = 14m/s \cr} $
Total velocity is $16\mathop i\limits^ \wedge + 14\mathop j\limits^ \wedge $

Speed is nothing but the magnitude of that velocity i.e
$\sqrt {{{16}^2} + {{14}^2}} = 21.26m/s$

Hence time taken to reach x coordinate of 16m is 2 seconds and y coordinate then is 24m and speed of the particle then is $21.26m/s$

Note: The reason why we solved the problem by splitting into x and y directions separately is because all the formulas we have are valid only for straight line motion and these formulas are valid only if acceleration is constant along the x and y axis and doesn’t vary with time.