Question

A particle starts from origin at $t = 0$ when a velocity $5.01\,m\,{s^{ - 1}}$ moves in x-y plane under the action of a force which produces a constant acceleration of $\left( {3.0\,\hat i + 2.0\,\hat j} \right)m\,{s^{ - 2}}$?
A. What is the y- coordinate of the particle at the instant its x-coordinate is $84m$ ?
B. What is the speed of the particle at this time?

Answer 1

Hint: Here, we will use the distance formula to calculate the y-coordinate at the instant of x-coordinate. Also, to calculate the speed of the particle we will calculate the velocity in case of both X-coordinate and y-coordinate and then we will take the magnitude of both the velocities. Remember here to calculate the time that the particle takes to displace from initial to final positions.

Complete step by step answer:
Here, we are given that the initial velocity of the particle in the x-coordinate is ${U_x} = 5.01m\,{s^{ - 1}}$.
And the initial velocity in the y-coordinate is ${U_y} = 0m\,{s^{ - 1}}$
Now, the acceleration in the x-coordinate is ${a_x} = 3\,m\,{s^{ - 2}}$
Also, the acceleration in the y-coordinate is ${a_y} = 2.0\,m\,{s^{ - 2}}$
$A.$ Now, the y-coordinate of the particle at the instant of x-coordinate can be calculated by using the distance formula of x-coordinate as given below
${\vec S_x} = {\vec U_x}t + \dfrac{1}{2}{\vec a_x}{t^2}$
Now, putting the values in the above equation, we get
$84\hat i = 5.01\,t\hat i + \dfrac{1}{2} \times 3 \times {t^2}\hat i$
$ \Rightarrow \,84\hat i = 5.01\,t\hat i + \dfrac{3}{2}\,{t^2}\hat i$
$ \Rightarrow \,168\hat i = \hat i\left( {10\,t + 3\,{t^2}} \right)$
$ \Rightarrow \,168 = 10t + 3{t^2}$
$ \Rightarrow \,3{t^2} + 10t - 168 = 0$
Now, factoring the above equation, we get
$3\,{t^2} + 28t - 18t - 168 = 0$
$ \Rightarrow \,t\left( {3t + 28} \right) - 6\left( {3t + 28} \right) = 0$
$ \Rightarrow \,\left( {3t + 28} \right)\left( {t - 6} \right) = 0$
$ \Rightarrow \,t = \dfrac{{ - 28}}{3},6$
Now, neglecting $t = \dfrac{{ - 28}}{3}$ , because time can never be negative.
$\therefore \,t = 6$
Now, at $t = 6$ , ${\vec S_x} = 84\hat i$
Now, the distance of y-coordinate is given by
${\vec S_y} = {\vec U_y}t + \dfrac{1}{2}{a_y}{t^2}$
Now, putting the values, we get
${\vec S_y} = 0 + \dfrac{1}{2} \times 2 \times 36$
$\therefore{\vec S_y} = 36m$

Therefore, the distance at y-coordinate at the instant is $36m$.

$B.$ Now, for calculating the velocity of the particle at this time, we will first calculate the velocity at x-coordinate as given by
${\vec V_x}\hat i = {\vec U_x}\hat i + {a_x}t\hat i$
$ \Rightarrow \,{\vec V_x} = 5.01\hat i + 3 \times 6\hat i$
$ \Rightarrow \,{\vec V_x} = 5\hat i + 18\hat i$
$ \Rightarrow \,{\vec V_x} = 23\hat i$
Now, the velocity at y-coordinate is given by
${\vec V_y} = {\vec U_y}\hat j + {\vec a_y}t$
$ \Rightarrow \,{\vec V_y} = 0 + 2 \times 6\hat j$
$ \Rightarrow \,{\vec V_y} = 12\hat j$
Now, the velocity at the time $t = 6$ is given by
$\left| V \right| = \sqrt {{{\vec V}_x} + {{\vec V}_y}} $
$ \Rightarrow \,\left| V \right| = \sqrt {{{\left( {23\hat i} \right)}^2} + {{\left( {12\hat j} \right)}^2}} $
$ \Rightarrow \,\left| V \right| = \sqrt {529 + 144} $
$ \Rightarrow \,\left| V \right| = \sqrt {673} $
$ \Rightarrow \,\left| V \right| = 25.94\,m{s^{ - 1}}$
$ \therefore \,V \simeq 26m{s^{ - 1}}$

Therefore, the velocity of the particle at $t = 6$ is $26m{s^{ - 1}}$.

Note:Here, it is important to calculate the velocity of the particle at x-coordinate and y-coordinate. The velocity of the particle is calculated by taking the magnitude of both the velocities. Also, it is important to calculate the time after reaching the final position, because time can never be zero.