Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 25 (Q.No. 15)
15.
Brake is applied on a vehicle which then skids a distance of 16 m before coming to stop. If the developed average coefficient of friction between the tyres and the pavement is 0.4, then the speed of the vehicle before skidding would have been nearly
20 kmph
30 kmph
40 kmph
50 kmph
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
6 comments Page 1 of 1.

Sachin said:   7 years ago
d =V^2/254f. Use this formula.
(2)

Bipin said:   9 years ago
Which formula used in this question.
(1)

Vivek said:   6 years ago
Sb = v^2/2gf.

V = √sb * 2gf,
V = √16 * 2 * 0.4 * 9.8.
V = 40kmph.
(1)

Shruti said:   9 years ago
Which formula we should use in this question?

Harsha said:   8 years ago
Braking dist = v^2/2gf here brak dist is given i.e. 16 m.

B.Singh said:   3 years ago
Sb = v^2/2gf.

V = √sb * 2gf,
V = √16 * 2 * 0.4 * 9.8.
V = 11.2 m/s,
V = 11.2 * 3.6 km/hr.
V = 40 km/hr (Approx).

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