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Importance of vision; acceleration numerical

Short answer

About 90 percent of driver information is visual; acuity, field, colour, glare, depth, and night vision affect safe driving. Given dv/dt=2-0.04v at 100 km/h, initial acceleration is 0.889 m/s2 and decreases, so 1.0 m/s2 is never reached afterward; it occurred 2.94 s earlier at 90 km/h.

Explain the importance of vision to Road users. The acceleration of a vehicle is given by the equation below: dv/dt = 2 − 0.04v. Compute the time when the acceleration of the vehicle will be 1.0 m/s² if the vehicle was initially travelling at 100 kph.

Importance of vision to road users

About 90 percent of the information a road user needs (road alignment, signs, signals, other vehicles, pedestrians and hazards) is received through the eyes, so vision is the most important sense in driving. Defective vision causes late detection, wrong judgement of speed and distance, and crashes.

  • Visual acuity decides the distance at which signs and objects can be read; sign letter heights are fixed to suit it.
  • Field of vision (clear cone about 3-5 degrees, fairly clear up to 10-12 degrees, peripheral up to about 120-160 degrees) fixes where signs are placed; it narrows with speed.
  • Colour vision is needed to read red, amber and green signals.
  • Glare vision and recovery affect night driving; they decide anti-glare design and lighting.
  • Depth perception is needed for judging gaps when overtaking and crossing.
  • Night vision needs street lighting, retro-reflective signs and markings.

Therefore design provides adequate sight distance (IRC:66, IRC:SP:23), legible signs (IRC:67), lighting and markings (IRC:35).

Numerical

Given dvdt=2−0.04v\dfrac{dv}{dt} = 2 - 0.04v (v in m/s), initial speed v0=100v_0 = 100 km/h =27.78= 27.78 m/s.

Since a=2−0.04va = 2 - 0.04v, differentiating gives dadt=−0.04 dvdt=−0.04a\dfrac{da}{dt} = -0.04\,\dfrac{dv}{dt} = -0.04a, so

a(t)=a0e−0.04t,a0=2−0.04(27.78)=0.889 m/s2a(t) = a_0 e^{-0.04t},\qquad a_0 = 2 - 0.04(27.78) = 0.889\ \text{m/s}^2

For a=1.0a = 1.0 m/s2^2 the speed would have to be v=(2−1)/0.04=25v = (2-1)/0.04 = 25 m/s =90= 90 km/h:

t=10.04ln⁡a01.0=25ln⁡(0.889)=−2.94 st = \frac{1}{0.04}\ln\frac{a_0}{1.0} = 25\ln(0.889) = -2.94\ \text{s}

Answer: the acceleration at the start (100 km/h) is only 0.889 m/s2^2, and it keeps falling as the vehicle speeds up, so it never reaches 1.0 m/s2^2 afterwards. The acceleration was 1.0 m/s2^2 about 2.94 s before the 100 km/h instant (t = -2.94 s, when the speed was 90 km/h). The value 1.0 m/s2^2 is reached after time zero only if the vehicle starts below 90 km/h (25 m/s); for example, from rest t=25ln⁡2=17.33t = 25\ln 2 = 17.33 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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