Question

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 $(8.0\hat{i}+2.0\hat{j})$ 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

Given:

The initial velocity of particle is $\overrightarrow{u}=10\hat{j}m/s$

The acceleration of the particle is $\overrightarrow{a}=8\hat{i}+2\hat{j}$

(a)

The position of the particle at any instant is given by:

$\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}{\overrightarrow{a}t}^2$

$x\hat{i}+y\hat{j}=4t^2\hat{i}+(10t+t^2)\hat{j}$

Comparing the X and Y component of the position:

$x=4t^2$

$x=16⟹t=2s$

$y=10t+t^2$

At t=2, $y=24m$

(b)

The speed of the particle at any instant is given by:

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$

$\overrightarrow{v}=10\hat{j}+\left(8\hat{i}+2\hat{j}\right)2$

$\overrightarrow{v}=14\hat{j}+16\hat{i}$

$|\overrightarrow{v}|=\sqrt{{14}^2+{16}^2}=21.26m/s$