Question

A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200m.What is the retardation of the car and how long does it take for the car to stop?​

Answer 1

Given :

A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200m.

To find :

• Retardation
• Time taken to stop a car

Solution :

★ Initial velocity (u) = 126 km/h

★ Final velocity (v) = 0

★ Distance travelled (s) = 200m

Convert km/h in m/s

→ Initial velocity = 126 km/h

→ u = 126 × 5/18

→ u = 7 × 5

→ u = 35 m/s

• According to the third equation of motion

→ v² = u² + 2as

• Substitute the values

→ (0)² = (35)² + 2 × a × 200

→ 0 = 1225 + 400a

→ 400a = - 1225

→ a = - 1225/400

→ a = - 3.06 m/s²

Focus Zone : Negative value shows a car in retardation

•°• Retardation of a car = 3.06m/s²

________________________________

• First equation of motion

→ v = u + at

→ 0 = 35 + (-3.06) × t

→ 0 = 35 - 3.06t

→ 3.06t = 35

→ t = 35/3.06

→ t = 11.4

•°• Time taken to stop a car = 11.4 sec

________________________________

Answer 2

$\huge{ \underbrace{ \text{Question}}}$

A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200m. What is the retardation of the car and how long does it take for the car to stop?

$\huge{ \underbrace{ \text{Answer}}}$

Given:-

$\sf{ \rightarrow{Initial \: velocity \: \bold{u} = 126kmh {}^{ - 1} = \frac{126 \times 1000}{3600} ms {}^{ - 1} = 35ms {}^{ - 1} }}$

$\sf{ \rightarrow{Final \: velocity \: \bold{v} = 0ms {}^{ - 1} }}$

$\sf{ \rightarrow{Distance \: \bold{s} = 200m}}$

To Find:-

$\sf{ \rightarrow{The \: retardation \: of \: the \: car}}$

$\sf{ \rightarrow{The \: time \: that \: will \: take \: to \: stop \: the \: car}}$

Solution:-

First, let's find the radiation of the car.

As we know that,

$\boxed{ \sf{ \blue{ {v}^{2} = {u}^{2} + 2as}}}$

According to the question,

$\sf{ {(0)}^{2} = (35 ) {}^{2} + 2 \times a \times 200}$

$\\ \sf{ \implies{0 = 1225 + 400 \times a}}$

$\\ \sf{ \implies{400 \times a = - 1225}}$

$\\ \sf{ \implies{a = \frac{ - 1225}{400} }}$

$\\ \sf{ \therefore{ \orange{a = - 3.0625}}}$

As the value of "a" is negative, the retardation is

$\sf{ - 3.0625 {ms}^{ - 2} }$

Now,

Let's find the time that took to stop the car.

As we know that,

$\boxed{ \sf{ \blue{v = u + at}}}$

According to the question,

$\sf{0 = 35 + ( - 3.0625) \times t}$

$\\ \sf{ \implies{ 35 - 3.0625 \times t}}$

$\\ \sf{ \implies{ 3.0625 \times t = 35}}$

$\\ \sf{ \implies{t = \frac{35}{3.0625} }}$

$\\ \sf{ \therefore{ \orange{t = 11.428}}}$

Hence, time taken to stop the car 11.428s

$\large{ \tt{ \underline{ \purple{Some \: important \: formulas:-}}}}$

$\sf{ \bold{1.}} \: a = \frac{v - u}{t} \\ \\ \sf{ \bold{2.}} \: s = (\frac{u + v}{2}) t \\ \\ \sf{ \bold{3.}} \: s = ut + \frac{1}{2} a {t}^{2} \\ \\ \sf{ \bold{4.}} \: {v}^{2} = {u}^{2} + 2as$