Kinematics-Examples I

17. Distance from home to school is 1.2 km. That distance is passed in 10 minutes. Calculate the average speed in m/s and km/s. 
$$\begin{align} & \begin{array}{*{35}{l}} s\text{ }=\text{ }1.2\text{ km }=\text{ }1.2\cdot 1000\text{ m }=\text{ }1200\text{ m} \\ \underline{t\text{ }=\text{ }10\text{ min }=\text{ }1\cdot 60\text{ s }=\text{ }60\text{ s}} \\ \end{array} \\ & v\text{ }=\text{ ?} \\ & v=\frac{s}{t}=\frac{1200}{60}=2\text{ m/s} \\ & v=2\text{ m/s} \\ & v\text{ }=\text{ 2 }\frac{{m}}{{s}}\cdot \frac{\frac{1\text{ km}}{1000\text{ }{\text{m}}}}{\frac{1\text{ h}}{3600\text{ }{\text{s}}}}=2\frac{3600}{1000}\frac{\text{km}}{\text{h}}=2\cdot 3.6\frac{\text{km}}{\text{h}}=7.2\text{ km/h} \\ & v=7.2\text{ km/h} \\ \end{align}$$
18. Speed of 15 m/s expressed in km/h is: 
$$\begin{align} & v=15\text{ m/s} \\ & v=1\text{5 }\frac{{m}}{{s}}\cdot \frac{\frac{1\text{ km}}{1000\text{ }{\text{m}}}}{\frac{1\text{ h}}{3600\text{ }{\text{s}}}}=15\frac{3600}{1000}\frac{\text{km}}{\text{h}}=15\cdot 3.6\text{ km/h}=54\text{ km/h} \\ & v=54\text{ km/h} \\ \end{align}$$
 19. In what time the body will travel the distance of 360 km, if the speed of the body is 10 m/s? 
$$\begin{align} & \begin{array}{*{35}{l}} s\text{ }=\text{ 360 km }=\text{ 360}\cdot 1000\text{ m }=\text{ 360000 m} \\ \underline{v\text{ }=\text{ }10\text{ m/s }} \\ \end{array} \\ & t\text{ }=\text{ ?} \\ & v={\frac{s}{t}}/{\cdot t}\;\Rightarrow s={vt}/{:v}\;\Rightarrow t=\frac{s}{v} \\ & t=\frac{s}{v}=\frac{360000}{10}=36000\text{ s} \\ & 1\text{ h}=3600\text{ s} \\ & t=36000\text{ }{\text{s}}\cdot \frac{1\text{ h}}{3600\text{ }{\text{s}}}=10\text{ h} \\ & t=10\text{ h} \\ \end{align}$$
  20. The speed of 600 m/min expressed in m/s is?
$$\begin{align} & v=600\text{ m/min} \\ & v=\text{600 }\frac{\text{m}}{{\min }}\cdot \frac{1{\min }}{60\text{ s}}=10\text{ m/s} \\ & v=10\text{ m/s} \\ \end{align}$$
  21. The speed of 54 km/h expressed in m/s is: 
$$\begin{align} & v=54\text{ km/h} \\ & v=54\text{ }\frac{{\text{km}}}{{\text{h}}}\cdot \frac{\frac{1000\text{ m}}{1\text{ }{\text{km}}}}{\frac{3600\text{ s}}{1\text{ }{\text{h}}}}=\frac{54}{3.6}\frac{\text{m}}{\text{s}}=15\text{ m/s} \\ & v=15\text{ m/s} \\ \end{align}$$
  22. In next figure the change of speed is shown in dependence of time: 

a) From previous figure determine what is the speed at moments  t₁=1s, t₂=2s, t₃=6s,

b) When was the body moving with constant speed? In what time

interval ? 

c) In what time interval is the speed of the body increasing and in 

what time interval is decreasing? 

$$\begin{align} & {{t}_{1}}=1\text{ s} \\ & {{t}_{2}}=3\text{ s} \\ & \underline{{{t}_{3}}=6\text{ s}} \\ & a){{v}_{1}}=4\text{ m/s}\text{, }{{\text{v}}_{2}}=8\text{ m/s}\text{, }{{\text{v}}_{3}}=8\text{ m/s} \\ & \text{b)t}\in \left[ 2,6 \right] \\ & c)\text{ Speed increasing in time interval t}\in \left[ 0,2 \right] \\ & \text{Speed decreasing in time interval t}\in \left[ 6,10 \right] \\ \end{align}$$ 23. Movement of the body can be expressed with the equation x = 5t where x is in meters and t is in seconds. 

a) Show the body movement on x,t graph and determine the speed of the moving body ?

The values of speed versus time are shown in the following table. 

t[s]	x[m]
0	0
1	5
2	10
3	15
4	20
5	25
6	30
7	35
8	40
9	45
10	50
11	55
12	60
13	65
14	70
15	75
16	80
17	85
18	90
19	95
20	100

Speed of the moving body can be calculated using the following 

$$v=\frac{s}{t}=\frac{5\cdot {t}}{{t}}=5\text{ m/s}$$

b) Determine where was the body when time t = 0 s. 

$$\begin{align} & x=5t+3 \\ & \underline{t=0\text{ s}} \\ & x=5\cdot 0+3=3\text{ m} \\ \end{align}$$

c) Determine where was the body when time t = 5 s. 

$$\begin{align} & x=5t+3 \\ & \underline{t=5\text{ s}} \\ & x=5\cdot 5+3=28\text{ m} \\ \end{align}$$