Question

A car moving along a straight highway with speed of 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Answer 1

Hii dear,

# Answer-
a = -3.06 m/s^2
t = 11.44 s.

# Given-
u = 126 km/h = 35 m/s
v = 0
s = 200 m

# Explaination-
Now let's assume,
a = retardation produced in the car
t = time taken to stop the car

From third kinematic equation,
v^2 = v^2 + 2as
(0)^2 = 35^2 + 2×a×200
a = –3.06 m/s^2

From first kinematic equation,
v = u + at
t = (v–u)/a
t = (0-35)/(-3.06)
t = 11.44 s

Hence, the car will have retardation -3.06 m/s^2 and will stop in 11.44 s.

Hope that helped you.

Answer 2

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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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