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denpristay [2]
3 years ago
9

an athlete runs 5.4 laps around a circular track that is 400.0 m long. If this takes 540 s, what is the average velocity of the

athlete? What is the average speed of the athlete over the complete distance of the run?
Physics
1 answer:
Slav-nsk [51]3 years ago
6 0

Answer:

Average velocity is 0.296 m/s.

Average speed is 4.0  m/s.

Explanation:

Given:

Distance of the circular track is, D=400.0\ m

Number of laps ran is, n=5.4

Time taken for the run is, t=540\ s

Now, total distance covered in 5.4 laps = D_T=D\times n=400\times 5.4=2160\ m

Also, since the path is a circle, the final position of the athlete after 5.4 laps will be 0.4 of 400 m ahead of the starting point.

Distance covered in 0.4 laps is, \textrm{Displacemet}=0.4\times 400=160\ m

Therefore, the displacement of the athlete will be 160 m as the athlete is 160 m ahead of the starting point and displacement depends on the initial and final points only.

Now, average velocity is given as:

v_{avg}=\frac{\textrm{Displacemet}}{t}=\frac{160}{540}=0.296\ m/s

Average speed is the ratio of total distance covered to total time taken.

So, average speed = \frac{D_T}{t}=\frac{2160}{540}=4\ m/s

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Given:

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                                F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

                                F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81  

                                F_weight = 19109.88 N

- The restoring force of a spring is given by:

                                F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

                                 F_weight = F_spring

                                 19109.88 = k*x

                                 x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

                                 x = 19109.88 / 160,000

                                 x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

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