Question

Derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motion in mutually perpendicular directions​

Answer 1

Answer:

Let the initial velocity of the particle be

$\vec u=u_x\hat i+u_y\hat j$

The acceleration of the particle be

$\vec a=a_x\hat i+a_y\hat j$

After time t if the velocity is $\vec v=v_x\hat i+v_y\hat j$ and the displacement is $\vec s$

Then

Acceleration = rate of change of velocity

$\vec a=\frac{\vec v-\vec u}{t}$

$\implies \vec a \times t=\vec v-\vec u$

$\implies \boxed{\vec v=\vec u+\vec a t}$        ............ (1)

Average velocity

$=\frac{\vec u+\vec v}{2}$

$\vec s=\frac{\vec u+\vec u+\vec a t}{2}\times t$

$\implies \vec s=\frac{2\vec u t+\vec a t^2}{2}$

$\implies \boxed{\vec s=\vec u t+\frac{1}{2}\vec a t^2}$   ............(2)

Taking the dot product of eq (1) with itself

$\vec v.\vec v=(\vec u+\vec a t).(\vec u+\vec a t)$

$\implies \vec v.\vec v=\vec u.\vec u+2\vec u.\vec a t+\vec a.\vec a t^2$

$\implies \vec v.\vec v=\vec u.\vec u+2\vec a.(\vec u t+\frac{1}{2}\vec a t^2)$

$\implies \boxed{\vec v.\vec v=\vec u.\vec u+2\vec a.\vec s}$  ........ (3)

(1), (2) and (3) are required equations of motion

From equation (1)

$\vec v=\vec u+\vec a t$

$v_x\hat i+v_y\hat j=(u_x\hat i+u_y\hat j)+(a_x\hat i+a_y\hat j) t$

$\implies v_x\hat i+v_y\hat j=(u_x\hat i+a_x\hat i t)+(u_y\hat j+a_y\hat j t$

Comparing the x and y component

$v_x=u_x+a_xt$

And

$v_y=u_y+a_yt$

From equation (2)

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

$s_x\hat i+s_y\hat j=(u_x\hat i+u_y\hat j) t+\frac{1}{2}(a_x\hat i+a_y\hat j) t^2$

$\implies s_x\hat i+s_y\hat j=(u_x t+\frac{1}{2}a_x t^2)\hat i+(u_y t+\frac{1}{2}a_y t^2)\hat j$

Comparing the x and y components

$s_x=u_x t+\frac{1}{2}a_x t^2$

$s_y=u_y t+\frac{1}{2}a_y t^2$

Similarly for equation (3) we can say that the motion can be resolved in two independents mutually perpendicular directions.

Hope this helps.

Answer 2

Let the initial velocity of the particle be

\vec u=u_x\hat i+u_y\hat j

u

=u

x

i

^

+u

y

j

^

The acceleration of the particle be

\vec a=a_x\hat i+a_y\hat j

a

=a

x

i

^

+a

y

j

^

After time t if the velocity is \vec v=v_x\hat i+v_y\hat j

v

=v

x

i

^

+v

y

j

^

and the displacement is \vec s

s

Then

Acceleration = rate of change of velocity

\vec a=\frac{\vec v-\vec u}{t}

a

=

t

v

−

u

\implies \vec a \times t=\vec v-\vec u⟹

a

×t=

v

−

u

\implies \boxed{\vec v=\vec u+\vec a t}⟹

v

=

u

+

a

t

............ (1)

Average velocity

=\frac{\vec u+\vec v}{2}=

2

u

+

v

\vec s=\frac{\vec u+\vec u+\vec a t}{2}\times t

s

=

2

u

+

u

+

a

t

×t

\implies \vec s=\frac{2\vec u t+\vec a t^2}{2}⟹

s

=

2

2

u

t+

a

t

2

\implies \boxed{\vec s=\vec u t+\frac{1}{2}\vec a t^2}⟹

s

=

u

t+

2

1

a

t

2

............(2)

Taking the dot product of eq (1) with itself

\vec v.\vec v=(\vec u+\vec a t).(\vec u+\vec a t)

v

.

v

=(

u

+

a

t).(

u

+

a

t)

\implies \vec v.\vec v=\vec u.\vec u+2\vec u.\vec a t+\vec a.\vec a t^2⟹

v

.

v

=

u

.

u

+2

u

.

a

t+

a

.

a

t

2

\implies \vec v.\vec v=\vec u.\vec u+2\vec a.(\vec u t+\frac{1}{2}\vec a t^2)⟹

v

.

v

=

u

.

u

+2

a

.(

u

t+

2

1

a

t

2

)

\implies \boxed{\vec v.\vec v=\vec u.\vec u+2\vec a.\vec s}⟹

v

.

v

=

u

.

u

+2

a

.

s

........ (3)

(1), (2) and (3) are required equations of motion

From equation (1)

\vec v=\vec u+\vec a t

v

=

u

+

a

t

v_x\hat i+v_y\hat j=(u_x\hat i+u_y\hat j)+(a_x\hat i+a_y\hat j) tv

x

i

^

+v

y

j

^

=(u

x

i

^

+u

y

j

^

)+(a

x

i

^

+a

y

j

^

)t

\implies v_x\hat i+v_y\hat j=(u_x\hat i+a_x\hat i t)+(u_y\hat j+a_y\hat j t⟹v

x

i

^

+v

y

j

^

=(u

x

i

^

+a

x

i

^

t)+(u

y

j

^

+a

y

j

^

t

Comparing the x and y component

v_x=u_x+a_xtv

x

=u

x

+a

x

t

And

v_y=u_y+a_ytv

y

=u

y

+a

y

t

From equation (2)

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

s

=

u

t+

2

1

a

t

2

s_x\hat i+s_y\hat j=(u_x\hat i+u_y\hat j) t+\frac{1}{2}(a_x\hat i+a_y\hat j) t^2s

x

i

^

+s

y

j

^

=(u

x

i

^

+u

y

j

^

)t+

2

1

(a

x

i

^

+a

y

j

^

)t

2

\implies s_x\hat i+s_y\hat j=(u_x t+\frac{1}{2}a_x t^2)\hat i+(u_y t+\frac{1}{2}a_y t^2)\hat j⟹s

x

i

^

+s

y

j

^

=(u

x

t+

2

1

a

x

t

2

)

i

^

+(u

y

t+

2

1

a

y

t

2

)

j

^

Comparing the x and y components

s_x=u_x t+\frac{1}{2}a_x t^2s

x

=u

x

t+

2

1

a

x

t

2

s_y=u_y t+\frac{1}{2}a_y t^2s

y

=u

y

t+

2

1

a mutually perpendicular directions.

Hope this helps.