Question

A particle is moving along a straight line with constant acceleration 4 metre per second square what will be its velocity and displacement at equals to 2 second if its initial velocity is 3 metre per second and initial displacement is 6 metre​

Answer 1

Initial Velocity = 3 m/sec

Initial Displacement = 6 m

Acceleration = 4 m/sec^2

$\rule{200}{2}$

The acceleration is constant.

Time = 2 second.

$\rule{200}{2}$

We have time,initial velocity and acceleration.

Applying 2nd Equation of motion:-

$s=ut+\dfrac{1}{2}at^2\\ \\ \\s=3*2+\dfrac{1}{2}4*4\\ \\ \\s=6+8\\ \\ \\s=14 m$

$\rule{200}{2}$

Hence:-

Total Displacement will be 14+6 = 20 m

$\rule{200}{2}$

From 1st Equation of Motion :-

$a=\dfrac{v-u}{t}\\ \\ \\a=\dfrac{v-3}{2}\\ \\ \\v-3=8\\ \\ \\v=8+3\\ \\ \\v=11\:m/sec$

$\rule{200}{2}$

Answer 2

Motion in a Straight Line :

Given, Acceleration, a = $\mathsf{4 \:m{s}^{-2}}$

Time, t = $\mathsf{2\:s}$

Initial velocity, u = $\mathsf{3 \:m{s}^{-1}}$

Initial displacement, $\mathsf{s_1}$ = $\mathsf{6\:m}$

Since, Final velocity is not given. So, we have only one choice to use "Newton's 2nd Equation of Motion".

$\boxed{\mathsf{\green{s\:=\:ut\:+\:{\dfrac{1}{2}\:*\:a{t}^{2}}}}}$

s = $\mathsf{ 3*2\:+\:{\dfrac{1}{2}\:*\:4*{2}^{2}}}$

s = $\mathsf{6\:+\:8}$

$\boxed{\mathsf{\green{Displacement,s\:=\:14 m.}}}$

Now, Total displacement = $\mathsf{s\:+\:s_1}$ = $\mathsf{14\:+\:6\:=\:20\:m}$

Using formula of Acceleration,

$\boxed{\mathsf{\green{a\:=\:{\dfrac{v\:-\:u}{t}}}}}$

$\mathsf{4\:=\:{\dfrac{v\:-\:3}{2}}}$

$\mathsf{8\:=\:v\:-\:3}$

$\mathsf{v\:=\:8\:+\:3}$

$\mathsf{v\:=\:11\:m{s}^{-1}}$

$\boxed{\mathsf{\green{Final\:velocity,v\:=\:11\: m{s}^{-1}}}}$