Question

solve in copy please don't post that i don't know ​

Answer 1

Explanation:

Given -

u = 25 m/s (initial velocity)

v = 0 m/s (final velocity)

t = 10 seconds (time)

To Find -

• The Distance travelled before the brakes are applied
• Retardation

As we know that :-

• s = (v + u/2)t

here,

s = distance travelled

v = final velocity

u = initial velocity

t = time

Then,

s = (0 + 25/2)×10

» 25 × 5

• » 125 m

Hence,

The Distance travelled before brakes is applied is 125 m

And

As we know that :-

• v = u + at

here,

v = final velocity

u = initial velocity

a = acceleration

t = time

Then,

» 0 = 25 + 10a

» -25 = 10a

» a = -25/10

» a = -2.5 m/s²

Hence,

The Retardation is -2.5 m/s²

Formula Used -

• v = u + at
• s = (v + u/2)t

here,

v = final velocity

u = initial velocity

a = acceleration

s = Distance travelled

t = time

Some related formulas :-

• s = ut + 1/2at²
• v² = u² + 2as

here,

v = final velocity

u = initial velocity

a = acceleration

t = time

s = distance travelled

Answer 2

GiveN :

• Initial velocity (u) = 25 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 10 s

To FinD :

• Retardation
• Distance travelled after force applied

SolutioN :

Use 1st equation of motion :

$\dashrightarrow {\boxed{\sf{v \: = \: u \: + \: at}}} \\ \\ \dashrightarrow \tt{0 \: = \: 25 \: + \: 10a} \\ \\ \dashrightarrow \tt{10a \: = \: -25} \\ \\ \dashrightarrow \tt{a \: = \: \dfrac{-25}{10}} \\ \\ \dashrightarrow \tt{a \: = \: -2.5} \\ \\ \underline{\sf{\therefore \: Retardation \: is \: - \: 2.5 \: ms^{-2}}}$

$\rule{200}{2}$

Now, use 3rd equation of motion :

$\dashrightarrow {\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \dashrightarrow \tt{0^2 \: - \: 25^2 \: = \: 2 \: \times \: -2.5 \: \times \: s} \\ \\ \dashrightarrow \tt{-625 \: = \: -5s} \\ \\ \dashrightarrow \tt{s \: = \: \dfrac{625}{5}} \\ \\ \dashrightarrow \tt{s \: = \: 125} \\ \\ \underline{\sf{\therefore \: Distance \: travelled \: after \: applying \: breaks \: is \: 125 \: m}}$