Question

On a highway with a
steep incline, the runaway truck ramp is constructed so that a vehicle that has
a lost its breaks can stop. The ramp is designed to slow a truck in succeeding
20-m distances by 10km/hr,
12km/hr, 14 km/hr If the ramp is 160 m long, will it stop a truck moving at 120 km/h when it reaches the ramp

Answer 1

This question is a simple mathematical problem in Arithmetic progression. It looks like a Physics and motion in a straight line problem.

The ramp of 160 meters is divided into 8 segments. First segment reduces speed by 10 kmph, second segment reduces speed by 12 kmph, the third segment reduces speed by further by 14 kmph and so on....

Reduction in speeds by each segment is as:  10, 12, 14, ..  8 terms 
     This is A. P. with a = 10 kmph and d = 2 kmph

     Reduction in speed by the n th segment = T_n = 10 + (n-1) 2 = 8 + 2 n   kmph
           n = 1, 2, .. 7 , 8

The speed of the truck at the bottom of the ramp be V kmph. 

The total reduction in speed at the end of n th segment of ramp is the total of the reductions by the previous segments.

       S_n = Σ T_n  for n = 1, 2, .., 8
              = Σ ( 8 + 2 n )  = 8 n + 2 n(n+1)/2 
              = 8 n + n² + n = n ( n + 9)
 
      S_n = 10, 22, 36 .. 

      S_8 = 8 (8+9) = 136 kmph
      S_7 = 7 (16) = 112 kmph 

If the truck starts to climb the ramp with an initial speed of 120 kmph, at the end of 7th segment, its speed will reduce to 120 - 112 = 8 kmph 

In the eight segment its speed will become 120 - 136 = -16 kmph.  So it starts sliding down.
 
So the truck with 120 kmph will stop on that ramp. The ramp can stop trucks with initial speed of 136 kmph too.

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Simple way:

Speed of truck at bottom of ramp = 120 kmph

Reduction by the 8 segments : 10 , 12, 14, 16, 18, 20 , 22, 24  kmph

Speed at the end of each segment:  
        =  120-10,   120-10-12,   120-10-12-14,  .... kmph
        =  110,   98,    84,  68,    50,   30,    8,  - 16  kmph

So in the eight segment it stops and slides down the ramp.