Question

A motor car slows down from 72km/hr to 36 km/he over a distance of 25 m.if the brakes are applied with the same force, calculate distance traveled by the car and total time in which the car comes to rest.​

Answer 1

Answer:

Explanation:

Given,

• Initial velocity of motor car, u = 72 km/h = 72 × 5/18 = 20 m/s
• Final velocity of motor car, v = 36 km/h = 36 × 5/18 = 10 m/s
• Distance covered by motor car is, s = 25 m

To Find,

• Distance covered, s
• Time taken, t

Formula to be used,

• 1st equation of motion, v = u + at
• 3rd equation of motion, v² - u² = 2as

Solution,

At 1st, we will find acceleration,

v² = u² + 2as

⇒ 10² = 20² + 2a × 25

⇒ 50a + 400 = 100

⇒ 50a = 100 - 400

⇒ 50a = -300

⇒ a = - 300/50

⇒ a = - 6 m/s²

Here, the acceleration is - 6 m/s².

Now, we will find out time taken,

v = u + at

⇒ 0 = 20 + (- 6) × t   (v = 0 m/s, as car stops)

⇒ - 6t = 20

⇒ t = 20/6

⇒ t = 3.33 seconds

Hence, the time taken is 3.33 seconds.

Now, the distance covered,

v² - u² = 2as

⇒ (0) - (20)² = 2 × (- 6) × s

⇒ 400 = -12s

⇒ 400/12 = s

⇒ s = 33.33 m.

Hence, the distance covered is 33.33 m.

Answer 2

Answer:

Given :-

• A motor car slows down from 72 km/hr to 36 km/hr over a distance of 25 m.
• The brakes are applied with the same force.

To Find :-

• What is the distance travelled by the car.
• What is the total time in which the car comes to rest.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Third Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken
• s = Distance Travelled

Solution :-

First, we have to convert km/h into m/s :

➲ Initial Velocity :

$\implies \sf Initial\: Velocity =\: 72\: km/h$

$\implies \sf Initial\: Velocity =\: 72 \times \dfrac{5}{18}\: m/s\: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Initial\: Velocity =\: \dfrac{\cancel{360}}{\cancel{18}}\: m/s$

$\implies \sf \bold{\purple{Initial\: Velocity =\: 20\: m/s}}$

➲ Final Velocity :

$\implies \sf Final\: Velocity =\: 36\: km/h$

$\implies \sf Final\: Velocity =\: 36 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\lgroup$

$\implies \sf Final\: Velocity =\: \dfrac{\cancel{180}}{\cancel{18}}\: m/s$

$\implies \sf\bold{\purple{Final\: Velocity =\: 10\: m/s}}$

Now, we have to find the acceleration :

Given :

• Initial Velocity = 20 m/s
• Final Velocity = 10 m/s
• Distance Travelled = 25 m

According to the question by using the formula we get,

$\implies \sf v^2 =\: u^2 + 2as$

$\implies \sf (10)^2 =\: (20)^2 + 2 \times a(25)$

$\implies \sf 100 =\: 400 + 2 \times 25a$

$\implies \sf 100 =\: 400 + 50a$

$\implies \sf 100 - 400 =\: 50a$

$\implies \sf - 300 =\: 50a$

$\implies \sf \dfrac{- \cancel{300}}{\cancel{50}} =\: a$

$\implies \sf - 6 =\: a$

$\implies \sf \bold{\green{a =\: - 6\: m/s^2}}$

Now, we have to find the distance travelled :

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 20 m/s
• Acceleration = - 6 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf (0)^2 =\: (20)^2 + 2 \times (- 6)s$

$\longrightarrow \sf 0 =\: 400 + (- 12)s$

$\longrightarrow \sf 0 - 400 =\: - 12s$

$\longrightarrow \sf {\cancel{-}} 400 =\: {\cancel{-}} 12s$

$\longrightarrow \sf 400 =\: 12s$

$\longrightarrow \sf \dfrac{400}{12} =\: s$

$\longrightarrow \sf 33.33 =\: s$

$\longrightarrow \sf\bold{\red{s =\: 33.33\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: travelled\: by\: the\: car\: is\: 33.33\: m\: .}}}}$

Now, we have to find the time taken :

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 20 m/s
• Acceleration = - 6 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf 0 =\: 20 + (- 6) \times t$

$\longrightarrow \sf 0 - 20 =\: - 6 \times t$

$\longrightarrow \sf {\cancel{-}} 20 =\: {\cancel{-}} 6 \times t$

$\longrightarrow \sf 20 =\: 6t$

$\longrightarrow \sf \dfrac{20}{6} =\: t$

$\longrightarrow \sf 3.33 =\: t$

$\longrightarrow \sf\bold{\red{t =\: 3.33\: seconds}}$

${\small{\bold{\underline{\therefore\: The \: total\: time\: taken\: in\: which\: the\: car\: comes\: to\: rest\: is\: 3.33\: seconds\: .}}}}$