1.

Calculating the stopping distance for a design speed of 75 kmph. Take the total reaction time 2.8 seconds and the coefficient of friction as 0.33.

Answer»

We have to find stopping distance, S
initial speed , u = 75 km/h = 75 × 5/18 = 125/6
= 20.83 m/s
total reaction time , t = 2.8 second.
before applying BREAK BODY in motion with constant speed.
so, distance TRAVELLED in reaction time, S' = ut
= 20.83 × 2.8 m = 58.32 m

now, after applying break, body start to decrease its speed with an acceleration.
because surface is ROUGH and coefficient of roughness is given 0.33
so, friction force = ma
umg = ma => a = ug = 0.33 × 10 = 3.3 m/s²

now, distance travelled by body after applying break , S" = u²/2ug = (20.83)²/2 × 3.3
= 65.74 m

so, stopping distance , S = S' + S"
= 58.32 + 65.74
= 124.06 m


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