Question

solve this fast pleasw its urgent​

Answer 1

Given :-

Initial velocity = 36 km /hr

In order to convert km /hr into m /s simply multiply by 5 /18

$\implies\mathtt{i = 36 \times \dfrac{5}{18}=10\dfrac{m}{s}}$

• initial velocity of the bus = 10 m /s

• Distance covered by the bus = 3 m

• Acceleration = 4 m /s ²

To find :-

• Final velocity

How to solve :-

• As the bus moves with uniform acceleration ,we can the third kinematic equation in order to find final velocity

Solution :-

As per the third kinematic equation ,

$\bigstar\pink{\boxed{\mathtt{ {v}^{2} = {u}^{2} + 2as}}}$

here ,

• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance

Now let's substitute the given values in the above equation ,

$\implies \mathtt{ {v}^{2} = 100 + 2 \times 4 \times 3}$

$\implies \mathtt{ {v}^{2} = 100 + 24}$

$\implies \mathtt{ {v}^{2} = 124}$

$\implies \mathtt{ \red{ v = 11.13 \dfrac{m}{s} }}$

The final velocity of the bus is 11 .13 m /s

Additional information :-

First equation of motion

$\bigstar\orange{\boxed{\mathtt{v = u + at}}}$

Second equation of motion

$\bigstar\purple{\boxed{\mathtt{s = ut + \dfrac{1}{2}a {t}^{2} }}}$

Distance traveled at n th second

$\bigstar\blue{\boxed{\mathtt{S_n= u + \dfrac{a}{2} (2n - 1)}}}$

Answer 2

$\huge\red Ans$

Given :-

Initial velocity = 36 km /hr

In order to convert km /hr into m /s simply multiply by 5 /18

\implies\mathtt{i = 36 \times \dfrac{5}{18}=10\dfrac{m}{s}}⟹i=36×

18

5

=10

s

m

initial velocity of the bus = 10 m /s

Distance covered by the bus = 3 m

Acceleration = 4 m /s ²

To find :-

Final velocity

How to solve :-

As the bus moves with uniform acceleration ,we can the third kinematic equation in order to find final velocity

Solution :-

As per the third kinematic equation ,

[\bigstar\pink{\boxed{\mathtt{ {v}^{2} = {u}^{2} + 2as}}}★

v

2

=u

2

+2as

here ,

v = final velocity

u = initial velocity

a = acceleration

s = distance

Now let's substitute the given values in the above equation ,

\implies \mathtt{ {v}^{2} = 100 + 2 \times 4 \times 3}⟹v

2

=100+2×4×3

\implies \mathtt{ {v}^{2} = 100 + 24}⟹v

2

=100+24

\implies \mathtt{ {v}^{2} = 124}⟹v

2

=124

\implies \mathtt{ \red{ v = 11.13 \dfrac{m}{s} }}⟹v=11.13

s

m

The final velocity of the bus is 11 .13 m /s

Additional information :-

First equation of motion

\bigstar\orange{\boxed{\mathtt{v = u + at}}}★

v=u+at

Second equation of motion

\bigstar\purple{\boxed{\mathtt{s = ut + \dfrac{1}{2}a {t}^{2} }}}★

s=ut+

2

1

at

2

Distance traveled at n th second

\bigstar\blue{\boxed{\mathtt{S_n= u + \dfrac{a}{2} (2n - 1)}}}★

S

n

=u+

2

a

(2n−1)