Question

A particle starts from the origin at t=0s with a velocity of 10$\widehat{j}\text{ }m{{s}^{-1}}$ and moves in the x-y plane with a constant acceleration of $\left( 8\widehat{i}+2\widehat{j} \right)m{{s}^{-2}}$.
(a) At what time is the x-coordinate of the particle is 16m? 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: Divide all the vectors into horizontal and vertical components. To find the time when x-coordinate is 16m and y-coordinate at this time use $s=ut+\dfrac{1}{2}a{{t}^{2}}$. For the speed of particles at this time use v=u+at for both the directions of motion.

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

Complete step by step answer:
(a) We will solve the given question by resolving the vectors into components.
It is given that the particle starts its motion at time t = 0 seconds. At time t = 0, the velocity of the particle is 10$\widehat{j}\text{ }m{{s}^{-1}}$. This means that the velocity of the particle is in upward direction, whose magnitude is 10 $m{{s}^{-1}}$.
It is given that the particle is moving with a constant acceleration, which is equal to $\overrightarrow{a}=\left( 8\widehat{i}+2\widehat{j} \right)m{{s}^{-2}}$.
Resolve the vector $\overrightarrow{a}$ in its horizontal and vertical components. Let the magnitudes of the horizontal and vertical components be ${{a}_{x}}$ and ${{a}_{y}}$ respectively.
Here, ${{a}_{x}}=8m{{s}^{-2}}$ and ${{a}_{y}}=2m{{s}^{-2}}$.
And the direction of ${{a}_{x}}$ is towards the right and the direction of ${{a}_{y}}$ is upwards.
It is asked to find the time when the x-coordinate of the particle is 16m. For this, we will use the kinematic equation, $s=ut+\dfrac{1}{2}a{{t}^{2}}$….. (i), for the horizontal motion of the particle.
Here, s is the displacement of the particle, u is the velocity in the considered direction at t=0, a is the acceleration in the considered direction and t is any instant of time.
It is given that the particle starts from the origin (i.e. x=0 and y=0). Therefore, when x-coordinate is 16m, s=16cm.
At t=0, the particle has no horizontal velocity. Hence, u=0.
And $a={{a}_{x}}=8m{{s}^{-1}}$.
Substitute the values in equation (i).
$\Rightarrow 16=(0)t+\dfrac{1}{2}(8){{t}^{2}}$
$\Rightarrow 16=4{{t}^{2}}$
$\Rightarrow t=\pm 2s$
t=-2s is discarded because time cannot be negative.
Hence, t=2s.
This means that after 2s the x-coordinate of the particle is 16m.
Let us calculate the y-coordinate at this time by using the same formula again for the vertical motion.
In this case, s=y and $u=10m{{s}^{-1}}$, t=2s and $a={{a}_{y}}=2m{{s}^{-1}}$.
This substituting the values in equation we get,
$\Rightarrow y=(10)(2)+\dfrac{1}{2}(2){{(2)}^{2}}=20+4=24m$.
Hence, at time 2s, the y-coordinate of the particle is 24cm.
(b) To calculate the speed of the particle at this instant time, let us first calculate the horizontal and vertical velocities of the particle at this time (${{v}_{x}}$ and ${{v}_{y}}$ respectively).
For this we will use the formula v=u+at for both the directions of motion.
For the horizontal motion, v=${{v}_{x}}$, u=0, $a={{a}_{x}}=8m{{s}^{-1}}$ and t=2s.
$\Rightarrow {{v}_{x}}=0+(8)(2)=16m{{s}^{-1}}$
For the vertical motion, v=${{v}_{y}}$, $u=10m{{s}^{-1}}$, $a={{a}_{y}}=2m{{s}^{-1}}$ and t=2s.
$\Rightarrow {{v}_{y}}=10+(2)(2)=10+4=14m{{s}^{-1}}$

The speed of the particle at this time will be $\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{16}^{2}}+{{14}^{2}}}=\sqrt{256+196}=\sqrt{452}=21.26m{{s}^{-1}}$

Note: Note that the speed of the particle at time 2 seconds is 21.26$m{{s}^{-1}}$. If the velocity of the particle was asked then this value will be the magnitude of the velocity.
The velocity of the particle at time t=2s is $\overrightarrow{v}=\left( 16\widehat{i}+14\widehat{j} \right)m{{s}^{-1}}$.