Answer:
The time taken is [tex]t_s = 31.16 s[/tex]
Explanation:
From the question we are told that
The length of the speed ramp is [tex]L = 104m[/tex]
The speed of the speed ramp is [tex]v_r = 2.1 m/s[/tex]
The time taken to cover the distance is [tex]t = 84 s[/tex]
The walking speed( when not on the speed ramp) required to cover the 104m is mathematically represented as
[tex]v_w = \frac{L}{t}[/tex]
Substituting values
[tex]v_w = \frac{104}{84}[/tex]
[tex]= 1.238 m/s[/tex]
The speed required to cover the 104m distance when on a speed speed ramp is mathematically represented as
[tex]v = The\ walking\ speed + speed \ of \ the \ speed \ ramp[/tex]
substituting values
[tex]v = 2.1 + 1.238[/tex]
[tex]v = 3.338 m/s[/tex]
Now the time taken to travel the 104m distance when walking on the speed ramp is mathematically evaluated as
[tex]t_s = \frac{L}{v}[/tex]
[tex]= \frac{104}{3.338}[/tex]
[tex]t_s = 31.16 s[/tex]
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
