A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured to the innermost vehicle path) is tentatively planned as 300 m. The limiting value for coefficient of side friction at 120 km/h is 0.09. What rate of superelevation is required for this curve at the 120 km/h design speed

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

Magnitude of the impulse = 75 Ns

Explanation:

Given:

Force exerts = 1500 N

Time = 50 ms = 0.05 s

Find:

Magnitude of the impulse

Computation:

Magnitude of the impulse = Force x Time

Magnitude of the impulse = 75 Ns