Question

1) a motorcycle running due west at a velocity of 72km/h is brought to rest in 2 seconds by the application of brakes. Distance covered by the motorcycle in these 2 seconds is

Answer 1

GIVEN :-

Initial velocity(u) = 72 km/h

Final velocity(v) = 0 m/s

(As the motorcycle is brought to rest)

Time(t) = 2 secs

TO CALCULATE :-

The distance covered by the motorcycle in 2 seconds.

FORMULAS TO BE USED :-

♠ v = u + at

• v is the final velocity.

• u is the initial velocity.

• a is the acceleration.

• t is the time taken.

$\rule{100}2$

♠ v² = u² + 2as

• v is the final velocity.

• u is the initial velocity.

• a is the acceleration.

• s is the distance covered.

SOLUTION :-

72 km/h = 72 × (5/18) = 5 × 4 = 20 m/s

→ v = u + at

⇒0 = 20 + a(2)

⇒-20 = 2a

⇒a = -20/2

⇒a = -10 m/s²

$\rule{90}2$

→ v² = u² + 2as

⇒(0)² = (20)² + 2(-10)s

⇒0 = 400 - 20s

⇒20s = 400

⇒s = 400/20

⇒s = 20 m

Therefore, the motorcycle covers a distance of 20 m in 2 secs.

$\underline{\rule{200}2}$

$\bigstar$ A related formula :-

s = ut + 1/2 at²

• s is the distance covered.

• u is the initial velocity.

• t is the time taken.

• a is the acceleration.

Answer 2

Given:

Initial velocity of motorcycle, u= 72 km/h

Final velocity of motorcycle, v= 0 km/h

(Since, Motorcycle comes to at rest)

Time taken by motorcycle, t= 2s

To Find:

Distance covered by the motorcycle in given time interval

Solution:

We know that,

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to third equation of motion for constant acceleration,

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{u=72\:km/h=72\times\dfrac{5}{18}\:m/s=20\:m/s}$

$\longrightarrow\rm{v=0\:km/h=0\times\dfrac{5}{18}\:m/s=0\:m/s}$

Let the acceleration of motorcycle be 'a'

On applying first equation of motion on given motorcycle, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=20+a(2)}$

$\longrightarrow\rm{-20=2a}$

$\longrightarrow\rm{2a=-20}$

$\longrightarrow\rm{a=\dfrac{-20}{2}}$

$\longrightarrow\rm{a=-10\:m/s^{2}}$

$\rule{190}{1}$

Let the distance covered by motorcycle be 's'

So, on applying third equation of motion on given motorcycle, we get

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{(0)^{2}-(20)^{2}=2(-10)s}$

$\longrightarrow\rm{-400=-20s}$

$\longrightarrow\rm{-20s=-400}$

$\longrightarrow\rm{s=\dfrac{\cancel{-400}}{\cancel{-20}}}$

$\longrightarrow\rm\green{s=20\:m}$

Hence, the distance covered by the motorcycle in given time interval is 20m.