Question 30 - Problems - Chapter 2

Problem:

The position of a person is given by s = 4.0t - 0.50t^2, where X is in meters and t is in seconds.

Question:

a) What is the average velocity between t=0 and t=8.0 s?

b) And between t=8.0 s and t=10.0 s?

Solution:

The position of the person is given by a parabolic position.

The factor of $t¨2$ is negative so the top of the parabol will be up.

The top (or maximal distance is given by the first derivative:

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

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

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

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

$$ \implies t = 4 $$

The average velocity:

$$ \overline v=\frac{\bigtriangleup s}{\bigtriangleup t} $$

The path in function of the time is described in the figure below:

from sympy import *
import numpy as np
import matplotlib.pyplot as plt
t_val = np.linspace(0,10,50)
s_val = 4*t_val - 0.5*t_val**2
plt.figure(figsize=(8,8))
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')
plt.hlines(y=0, xmin=-0.2, xmax=10.5, color='r', linestyles='dashed')
plt.hlines(y=8, xmin=-0.2, xmax=5, color='r', linestyles='dashed')
plt.annotate('Point of return = 8m', xy=(-0.5, 8.5))
plt.vlines(x=0, ymin=-10.2, ymax=8.5, color='r', linestyles='dashed')
plt.vlines(x=8, ymin=-10.2, ymax=8.5, color='r', linestyles='dashed')
plt.vlines(x=10, ymin=-10.2, ymax=8.5, color='r', linestyles='dashed')
plt.annotate('Start = 0s', xy=(-0.5, -11))
plt.annotate('Point = 8s', xy=(7.5, -11))
plt.annotate('End = 10s', xy=(9.5, -11))
plt.show()

Displacement $\Delta d_{0-8s}= 8 + (- 8) = 0m$

Traveled way $s_{0-8s} = |8| + |-8| = 16m$

Displacement $\Delta d_{8-10s}= -10m$

Traveled way $s_{8-10s} = |-10| = 10m$

t_8 = 8
t_10 = 2
d_8 = 0
s_8 = 16
d_10 = -10
s_10 = 10
v_v_8 = d_8/t_8
v_s_8 = s_8/t_8
v_v_10 = d_10/t_10
v_s_10 = s_10/t_10
print(f'a) The average velocity between 0s and 8s is: {v_v_8:.2f} m/s')
print(f'b) The average velocity between 8s and 10s is: {v_v_10:.2f} m/s')
print('')
print(f'- The average speed of the first part is {v_s_8:.2f} m/s an of the second part is {v_s_10:.2f} m/s')

a) The average velocity between 0s and 8s is: 0.00 m/s
b) The average velocity between 8s and 10s is: -5.00 m/s

- The average speed of the first part is 2.00 m/s an of the second part is 5.00 m/s