Question 23 - Problems - Chapter 2

Problem:

The position of a runner is given by $x = 4.0t-0.50t^2$, where x is in meters and t is in seconds.

Question:

What is the average speed between t = 0 and t = 8.0 s?

(Hint: Find the maximum value of x to determine each of the outward and backward distances.)

Solution:

The runner's trajectory is described by a parabola. He thus travels back and forth in distance.

The factor of t squared is negative so the top of the parabola points upwards.

We can determine the furthest distance the runner travels when the derivative of the distance is zero.

$$ \frac{dx}{dt}=\frac{d}{dt}(-0.5t^2+4t) $$

We find the turn aroud point by solving for speed = 0

$$ \implies-0.5x2t+4=0 $$

$$ \implies t=\frac{-4}{-1} $$

$$ \implies t = 4 $$

We could also let Python do the derivative using the Sympy (Symbolic Python) library. by mean of excersice we will do this here.

from sympy import *
import numpy as np
import matplotlib.pyplot as plt
t = symbols('t')
x = diff((4*t - 0.5*(t**2)),t,1)   # solve the derivative into t
print(f'The derivative is ({x})')
s = solve(x,t) # here We solve the result in t (for v = 0 the moment of turn around)
               # there is only one solution so we use element 0 of our solutions
print(f'The turn arround point is at {s[0]:.2f} seonds')

The derivative is (-1.0*t + 4)
The turn arround point is at 4.00 seonds

The distance run after the time he reached the turn around point is:

$$ s_{half} = \Big|4.(4s) - 0.5.(4s)^2\Big| $$

s_half = np.abs(4*4-1.5*4**2)
Time = 8 #s
print(f'The turn around point is at {s_half:.1f} m so the total distance is {2*s_half:.1f} m')
v_mean = (2 * s_half)/Time
print(f'The average speed over 8 seconds is {v_mean:.2f} m/s')

The turn around point is at 8.0 m so the total distance is 16.0 m
The average speed over 8 seconds is 2.00 m/s

t_val = np.linspace(0,8,100)
s_val = 4*t_val - 0.5*t_val**2
plt.figure(figsize=(8,6))
plt.plot(t_val, s_val)
plt.grid()
plt.title('The runners position in function of time')
plt.xlabel('t - time')
plt.ylabel('s - position')

Text(0,0.5,'s - position')