Question

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

Answer 1

Given:-

→ Initial velocity of the car = 126km/h

→ Distance covered = 200m

To find:-

→ Retardation of the car

→ Time taken by the car to stop

Solution:-

Firstly, let's convert the initial velocity of the car from km/h to m/s.

=> 1km/h = 5/18m/s

=> 126km/h = 5/18×126

=> 35m/s

In this case :-

• Final velocity (v) of the car will be zero,

as it ultimately stops .

By using the 3rd equation of motion, we get:-

=> v² - u² = 2as

=> 0 - (35)² = 2×a×200

=> -1225 = 400a

=> a = -1225/400

=> a = -3.0625 m/s²

[ Here, -ve sign shows retardation ]

Now, by using the 1st equation of motion, we get:-

=> v = u + at

=> 0 = 35 + (-3.0625)t

=> -35 = -3.0625t

=> t = -35/-3.0625

=> t = 11.42 s. [approximately]

Thus:-

• Retardation of the car is 3.0625 m/s².

• Time taken by the car to stop is 11.42 s.

Answer 2

Good question with Good Answer ◉‿◉

$\: \: \: \: \: \: \: \: \: \: \: u = 126 \: km \: h^{ - 1} \: = 35 \: m \: s^{ - 1} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:v \: = 0 \: ms^{ - 1} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a \: = \: ? \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: t \: = \: ? \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s \: = \: 200$

Now,

$v ^{2} - u ^{2} = 2as \\ 0 - (35) ^{2} = 2 \times a \times 200 \\ - 1225 = 400 \times a \\ \\ a = \frac{ - 1225}{400} \\ \\ = \frac{49}{16}$

and,

$v = u + at \\ 0 = 35 - \frac{49}{16} t \\ \\ t = \frac{ - 35 \times 16}{ - 49} \\ \\ t = \frac{80}{7}$

$\LARGE\therefore$ a = 3.06 m/s^2

and, t = 11.4 sec