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 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')
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')
