Question

a moving car along a straight line highway with speed of 126/hrs 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

Answer :-

→ Retardation is 3.0625 m/s² .

→ Time taken to stop is 11.42 seconds .

Explanation :-

We have :-

→ Initial velocity (u) = 126 km/h

→ Stopping distance (s) = 200 m

→ Final velocity (v) = 0 m/s

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Firstly, let's convert the unit of initial velocity (u) from km/h to m/s .

⇒ 1 km/h = 5/18 m/s

⇒ 126 km/h = 5/18(126)

⇒ 35 m/s

Now, let's calculate the retardation [-ve acceleration] of the car by using the 3rd equation of motion .

v² - u² = 2as

⇒ 0 - (35)² = 2(a)(200)

⇒ -1225 = 400a

⇒ a = -1225/400

⇒ a = -3.0625 m/s²

[Here, -ve sign shows retardation] .

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Finally, let's calculate the required time taken by using the 1st equation of motion .

v = u + at

⇒ 0 = 35 + (-3.0625)t

⇒ 0 - 35 = -3.0625t

⇒ -35 = -3.0625t

⇒ t = -35/-3.0625

⇒ t ≈ 11.42 s

Answer 2

Given :-

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

To Find :-

Retardation

Time

Solution :-

We know that

1 km/h = 5/18 m/s

126 km/h = 126 × 5/18

126 km/h = 7 × 5

126 km/h = 35 m/s

(v)² = (u)² + 2as

(0)² = (35)² + 2(a)(200)

(0 × 0) = (35 × 35) + 400a

0 = 1225 + 400a

0 - 1225 = 400a

-1225 = 400a

-1225/400 = a

-49/16 = a

Now

v = u + at

0 = (35) + (-49/16)(t)

0 = 35 - 49t/16

0 = 560 - 49t/16

(0)(16) = 560 - 49t

0 = 560 - 49t

0 - 560 = -49t

-560 = -49t

-560/-49 = t

80/7 = t

Hence

Acceleration is -49/16 m/s² and time 80/7 second