Question

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

Answer 1

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Initial velocity of the car, u = 126 km/h = 35 m/s

Final velocity of the car, v = 0

Distance covered by the car before coming to rest, s = 200 m

Retardation produced in the car = a

From third equation of motion, a can be calculated as:

V^2 - U^2 = 2aS

(0)^2 - (35)^2 = 2 . a . 200

a = ( 35×35)/(2× 200)

a= 3.06 m/ s^2

From first equation of motion, time (t) taken by the car to stop can be obtained as:

V = U + at

t = (V- U )/a = (-35)/(-3.06) = 11.44 sec

I hope, this will help you

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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