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katrin2010 [14]

3 years ago

11

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

on which you can either stand or walk. Suppose a speed ramp has a length of 104 m and is moving at a speed of 2.1 m/s relative to the ground. In addition, suppose you can cover this distance in 84 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 104 m using the speed ramp

2 answers:

mezya [45]3 years ago

8 0

Answer:

Time t = 31.16s

Therefore, it takes 31.16 seconds to travel 104 m using the speed ramp

Explanation:

Length of speed ramp = 104 m

Time taken to cover the length walking = 84s

Walking speed = length/time = 104/84 = 1.238 m/s

Speed ramp speed relative to the ground = 2.1m/s

Total speed of combined speed ramp and walking on speed ramp is;

Vt = 2.1 + 1.238 = 3.338 m/s

Time taken to cover 104 m with combined speed ramp and walking on speed ramp;

Time t = distance/speed = 104/3.338

Time t = 31.16s

Therefore, it takes 31.16 seconds to travel 104 m using the speed ramp

Amanda [17]3 years ago

4 0

Answer:

The time taken is $t_s = 31.16 s$

Explanation:

From the question we are told that

The length of the speed ramp is $L = 104m$

The speed of the speed ramp is $v_r = 2.1 m/s$

The time taken to cover the distance is $t = 84 s$

The walking speed( when not on the speed ramp) required to cover the 104m is mathematically represented as

$v_w = \frac{L}{t}$

Substituting values

$v_w = \frac{104}{84}$

$= 1.238 m/s$

The speed required to cover the 104m distance when on a speed speed ramp is mathematically represented as

$v = The\ walking\ speed + speed \ of \ the \ speed \ ramp$

substituting values

$v = 2.1 + 1.238$

$v = 3.338 m/s$

Now the time taken to travel the 104m distance when walking on the speed ramp is mathematically evaluated as

$t_s = \frac{L}{v}$

$= \frac{104}{3.338}$

$t_s = 31.16 s$

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