Please provide formula and detailed workings. Thank you. This is a mathematical

Question

Please provide formula and detailed workings. Thank you.

This is a mathematical modeling question on speed humps. You are working as a transportation engineer with the Land Transport Authority (LTA). You will have to use your mathematical modeling skills to advise LTA on how far apart should two successive speed humps, in metres (m), be spaced along a stretch of road whose speed limit is 40 km hr^-1. The speed with which a vehicle crosses a speed hump is 2.84 m s^-1. The vehicle's acceleration while leaving speed hump is 1.61 m s^-2, while its acceleration when braking on approach to a speed hump is -6.43 m S^-2. This negative acceleration is a deceleration. Calculate the distance, in m, between two speed humps, giving your answer to 3 decimal places

Explanation / Answer

Speed Limit =40km/hr = 40*1000/3600 = 11.11 m/s

V = 2.84m/s ; a =1.61m/s2 ;

and then deacceleration = 6.43 m/s2

For Acceleration Part

At t= 0 vehicle is on first bump with speed(u) = 2.84 m/s

then with acceleration(a) = 1.61 m/s2 it reached final velocity(v) = 11.11m/s

Distance travelled to reach final velocity (D1) = (v2 - u2)/2a = (11.11*11.11 - 2.84*2.84)/3.22 =32.8281 m

For Decelertion Part

initial velocity for this case (u) =11.11m/s

final velocity (v) = 2.84m/s

decelearation (a) = -6.43 m/s2

distance = (v2 - u2)/2a = -(11.11*11.11 - 2.84*2.84)/2(-6.43) = 8.9709 m

Total distance = 32.8281 m +  8.9709 m =41.799m

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