91Ó°ÊÓ

It is given that,

A constant acceleration of pebble in xy plane is

$\mathrm{a}⃗=(5.00\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{i}}+(7.00\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}}.$

Velocity of pebble at$\mathrm{t}=0\mathrm{s}$

role="math" localid="1656418239213" $\mathrm{v}⃗=(4.0 \mathrm{m}/\mathrm{s})\hat{\mathrm{i}}$

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Using these kinematic equations time to travel the distanceof 12mcan be found. Further, using that time in first kinematic equation, the velocity of pebbles and its magnitude along x axis and the angle between vectorand +x axis will be computed.It is possible to find the magnitude of given velocity and angle made by the velocity with particular axis using the concept of projectile motion and vector algebra.

The displacement in kinematic equation can be written as,

$\mathrm{x}={\mathrm{v}}_0\mathrm{t}+1/2{\mathrm{at}}^2\left(\mathrm{i}\right)$

Where, ${\mathrm{v}}_0$ is the initial velocity

The final velocity is given by

$\mathrm{v}={\mathrm{v}}_0+\mathrm{at}\left(\mathrm{ii}\right)$

The magnitude of the velocity can be written as,

$\mathrm{v}=|\mathrm{v}⃗|=\sqrt{({\mathrm{v}}_{\mathrm{x}}{)}^2+({\mathrm{v}}_{\mathrm{y}}{)}^2}\left(\mathrm{iii}\right)$

The direction of the velocity is given by,

localid="1656418669599" $\mathrm{θ}={\tan }^{-1}\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}\left(\mathrm{iv}\right)$

For distance of 12m, pebbles need a time

$$\begin{gathered} 12.0 \mathrm{m}=\left(4.0\frac{\mathrm{m}}{\mathrm{s}}\right)×\mathrm{t}+\frac{1}{2}\left(5.00\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)×{\mathrm{t}}^2 \\ \mathrm{t}=1.53 \mathrm{s} \end{gathered}$$

Substituting the above calculated time in equation (ii), velocity of pebble is,

$$\begin{gathered} \mathrm{v}⃗=(4.0 \mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+\left[\left(5.0\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\hat{\mathrm{i}}+\left(7.0\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\hat{\mathrm{j}}\right]×(1.53\mathrm{s}{)}^2 \\ \mathrm{v}⃗=(11.7\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+(10.7\mathrm{m}/\mathrm{s})\hat{\mathrm{j}} \end{gathered}$$$v⃗=(4.0 m/s)\hat{i}+\left[\left(5.0\frac{m}{s^2}\right)\hat{i}+\left(7.0\frac{m}{s^2}\right)\hat{j}\right]×(1.53s{)}^{2}v⃗=(11.7"m/s")i^+(10.7"m/s"\left)\right("j")^$

Therefore, using equation (iii) the magnitude of the velocity is,

$$\begin{gathered} |\mathrm{v}⃗|={\left(11.7\frac{\mathrm{m}}{\mathrm{s}}\right)}^2+{\left(10.7\frac{\mathrm{m}}{\mathrm{s}}\right)}^2 \\ \mathrm{v}=15.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the magnitude of the velocity is 15.8 m/s.

Using equation (iv), theangle of pebble’s velocity is,

${\tan }^{-1}\left(\frac{10.7}{11.7}\right)=42.6^{\mathrm{o}}$

Hence,the angle of pebble’s velocity is $42.6^{∘}$