Question

A particle starts from the origin at $t = 0s$ with a velocity of $10.0\hat jm/s$ and moves in the x-y plane with a constant acceleration of $(8.0\hat i + 2.0\hat j)m/{s^2}$
a) At what time is the x- coordinate of the particle $16m?$ 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 : To solve this question we have to know about velocity and acceleration properly. We know that velocity is a proportion of the measure of work a Team can handle during a solitary Sprint and is the critical measurement in Scrum. And also we know that, In mechanics, speed increase is the pace of progress of the speed of an item regarding time. Speed increases are vector amounts (in that they have greatness and course).

Complete Step By Step Answer:
According to the question we can say, the initial velocity of the particle is $\vec u = 10.0\hat jm/s$
And, the acceleration of the particle is $\vec a = (8.0\hat i + 2.0\hat j)$
We know that, the position of the particle at any instant is,
$\vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}$
Now, putting the values of acceleration and initial velocity we will get,
$x\hat i + y\hat j = 4{t^2}\hat i + (10t + {t^2})\hat j$
Now, comparing the X and Y component of the position we can say,
$\begin{gathered} x = 4{t^2} \\\ x = 16 \Rightarrow t = 2s \\\ \end{gathered}$
And, $y = 10t + {t^2}$
At $t = 2$
$y = 24m$
Now, we know that the speed of the particle at any instant is,
$\vec v = \vec u + \vec at$
Now, after putting the values we can write,
$\vec v = 10\hat j + (8\hat i + 2\hat j)2$
$\vec v = 14\hat j + 16\hat i$
$\left| {\vec v} \right| = \sqrt {{{14}^2} + {{16}^2}} = 21.26m/s$

Note :
We also have to know about the units of the speed or the velocity and the acceleration of any moving body. We know that, The SI unit of velocity is meter per second (m/s). On the other hand, the speed extent can likewise be communicated in centimeters each second (cm/s).