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
Discussion:
6 comments Page 1 of 1.
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).
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).
Vivek said:
6 years ago
Sb = v^2/2gf.
V = √sb * 2gf,
V = √16 * 2 * 0.4 * 9.8.
V = 40kmph.
V = √sb * 2gf,
V = √16 * 2 * 0.4 * 9.8.
V = 40kmph.
(1)
Sachin said:
7 years ago
d =V^2/254f. Use this formula.
(2)
Harsha said:
8 years ago
Braking dist = v^2/2gf here brak dist is given i.e. 16 m.
Bipin said:
9 years ago
Which formula used in this question.
(1)
Shruti said:
9 years ago
Which formula we should use in this question?
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