Question:
While driving at a speed of \(30\) kmph (with available friction \(0.4\)) down the grade, the driver requires a braking distance twice that required for stopping the vehicle when he travels up the same grade. The grade is
While driving at a speed of \(30\) kmph (with available friction \(0.4\)) down the grade, the driver requires a braking distance twice that required for stopping the vehicle when he travels up the same grade. The grade is
Show Hint
Braking distance on grades:
\[
\boxed{
BD=\frac{V^2}{254(f-G)}
}
\]
for downgrades, and
\[
\boxed{
BD=\frac{V^2}{254(f+G)}
}
\]
for upgrades.
Updated On: Jul 23, 2026
- \(7\%\)
- \(10.6\%\)
- \(13.3\%\)
- \(33.3\%\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
The braking distance on a grade is
\[
\boxed{
BD=\frac{V^2}{254(f\pm G)}
}
\]
where
\[
f=\text{Coefficient of longitudinal friction},
\]
\[
G=\text{Grade (decimal)},
\]
\[
(+)=\text{Upgrade},
\]
\[
(-)=\text{Downgrade}.
\]
Step 1: Use the given condition. The downhill braking distance is twice the uphill braking distance. Hence, \[ \frac{1}{f-G} = \frac{2}{f+G} \]
Step 2: Substitute \(f=0.4\). \[ 0.4+G = 2(0.4-G) \] \[ 0.4+G = 0.8-2G \] \[ 3G = 0.4 \] \[ G = 0.133 \] Thus, \[ \boxed{ G=13.3\% } \] Therefore, the correct option is \[ \boxed{(C)\;13.3\%.} \]
Step 1: Use the given condition. The downhill braking distance is twice the uphill braking distance. Hence, \[ \frac{1}{f-G} = \frac{2}{f+G} \]
Step 2: Substitute \(f=0.4\). \[ 0.4+G = 2(0.4-G) \] \[ 0.4+G = 0.8-2G \] \[ 3G = 0.4 \] \[ G = 0.133 \] Thus, \[ \boxed{ G=13.3\% } \] Therefore, the correct option is \[ \boxed{(C)\;13.3\%.} \]
Was this answer helpful?
0
0
Top TS PGECET Civil Engineering Questions
- A man weighing \( W \) newtons entered a lift which moves with an acceleration of \( a \, \text{m/s}^2 \). When the lift is moving downward, the force exerted by the man on the floor of the lift is:
- The magnitude of the resultant of two equal forces each of magnitude \( F \) is \( \sqrt{2}F \). Then the angle between their line of action is:
- The equation of motion of a particle starting from rest along a straight line is \( x = t^3 - 3t^2 + 5 \). The ratio of the accelerations after 5 s and 3 s will be:
- A tennis player returns a ball by hitting it with a racket. If the force exerted by the racket on the ball is increased, then the impulse imparted to the ball:
- Radius of gyration is that distance which when squared and multiplied with total mass of the body gives
View More Questions
Top TS PGECET Highway Engineering Questions
- The design of horizontal and vertical alignments, super elevation and gradient is worst affected by
- The ratio of passenger car unit of pedal cycle to passenger car unit of car is
- If the super elevation of the highway provided is zero, then the design speed of highway having a curve of 200 m and coefficient of friction 0.10 is
- The best direction of a runway is along the direction of
- Which of the below is not a type of force acting on the cement concrete pavement?
View More Questions
Top TS PGECET Questions
- If \( P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \) is the modal matrix of \( A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \) then the sum of all the elements of \( P^{-1}AP \) is
- If the system of equations \( kx + y + z = k - 1 \), \( x + ky + z = k - 1 \), and \( x + y + kz = k - 1 \) has infinite solutions, then \( k \) is
- The value of the integral \( \int_C (2xy - x^2) \, dx + (x^2 + y^2) \, dy \) where \( C \) is the boundary of the region enclosed by \( y = x^2 \) and \( y^2 = x \), described in the positive sense, is
- Let \( f(x) = \log x \). The number \( C \) strictly between \( e^2 \) and \( e^3 \) such that its reciprocal is equal to \( \frac{f(e^3) - f(e^2)}{e^3 - e^2} \) is
- The solution of the differential equation \( (D^2 + 2)y = x^2 \) is
View More Questions