Solved Problem on Kinematics
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The motion of a car, which moves with constant speed, is described by the following table

 t (h) 1 2 4 7 9 11 12 S (km) 100 200 450 600 400 200 100

from the table data find:
a) The average speed of the car between the instants 1 h and 2 h;
b) The average speed of the car between the instants 4 h and 7 h;
c) The average speed of the car between the instants 9 h and 12 h;
d) The average speed of the car between the instants 1 and 12 h.

Problem diagram

figure 1

We choose a reference frame oriented to the right.

Solution

a) Using the expression for the average speed
$\bbox[#99CCFF,10px] {v_{\text{m}}=\frac{\Delta S}{\Delta t}=\frac{S_{\text{f}}-S_{\text{i}}}{t_{\text {f}}-t_{\text{i}}}} \tag{I}$
$v_{\text{m}}=\frac{200-100}{2-1}\\ v_{\text{m}}=\frac{100}{1}$
$\bbox[#FFCCCC,10px] {v_{\text{m}}=100\;\text{km/h}}$

b) Using the expression (I) for the average speed
$v_{\text{m}}=\frac{600-450}{7-4}\\ v_{\text{m}}=\frac{150}{3}$
$\bbox[#FFCCCC,10px] {v_{\text{m}}=50\;\text{km/h}}$

c) Using the expression (I) for the average speed
$v_{\text{m}}=\frac{100-400}{12-9}\\ v_{\text{m}}=-\frac{300}{3}$
$\bbox[#FFCCCC,10px] {v_{\text{m}}=-100\;\text{km/h}}$

d) Using the expression (I) for the average speed
$v_{\text{m}}\;=\;\frac{100-100}{12-1}\\ v_{\text{m}}\;=\;\frac{0}{11}$
$\bbox[#FFCCCC,10px] {v_{\text{m}}=0}$

Note: The average zero does not mean that the car remained at rest. If the average speed is zero, the object may have remained at rest in the same position all the time or may have moved and returned to the same position (as happened here). The average speed gives an overview of the change of speed but does not indicate the point variations that occurred during the movement.