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>> $Basic Introduction to Veloc...$

Basic Introduction to Velocity

Q: A man walks on a straight road from his home to a market \$2.5\ km\$ away with a speed of \$5\ km\ h^{-1}\$ . Finding the market closed, he instantly turns and walks back home with a speed of \$7.5\ km\ h^{-1}\$. What is the (a) magnitude of average velocity, and(b) average speed of the man over the interval of time (i) \$0\$ to \$30\ min\$, (ii) \$0\$ to \$50\ min\$ (iii) \$0\$ to \$40\ min\$?

Distance to market \$s=2.5 km=2.5 \times 10^3=2500 m\$Speed with which he goes to market \$=5 km/h=5 \dfrac{10^3}{3600}=\dfrac{25}{18} m/s\$Speed with which he comes back \$=7.5 km/h=7.5 \times \dfrac{10^3}{3600}=\dfrac{75}{36} m/s\$(a)Average velocity is zero since his displacement is zero.(b) (i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market:\$\dfrac{2.5}{5}=1/2 h=30\$ minutes.Average speed over this interval \$=5 km/h\$(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :\$7.5 \times \dfrac{1}{3}=2.5 km\$His average speed in 0 to 50 minutes: \$V_{avg}=\dfrac{distance \quad traveled}{time}\$\$=\dfrac{2.5+2.5}{(50/60)}=6 km/h\$(iii)In 40-30=10 minutes he travels a distance of :\$7.5 \times \dfrac{1}{6}=1.25 km\$\$V_{avg}=\dfrac{2.5+1.25}{(40/60)}=5.625 km/h\$

Q: A car moving on a straight road covers one third of a certain distance with 20 Km/h and the rest with 60 Km/h. The average speed is :

Given,Velocity for one third of distance \$ {{V}_{1}}=20\,km/h \$ Velocity for two third of distance \$ {{V}_{2}}=60\,km/h \$ Time take to cover one third of distance \$ {{t}_{1}}=\dfrac{d/3}{{{V}_{1}}}=\dfrac{d/3}{20} \$ Time taken to cover two third of distance \$ {{t}_{2}}=\dfrac{2d/3}{{{V}_{2}}}=\dfrac{2d/3}{60} \$ \$ \text{Average velocity =}\dfrac{\text{Total}\,\text{distance}\,}{\text{total}\,\text{time}} \$ \${{V}_{av}}=\dfrac{d}{{{t}_{1}}+{{t}_{2}}}=\dfrac{d}{\dfrac{d/3}{20}+\dfrac{2d/3}{60}}=36\,km/h\$

Q: A particle moving in a straight line covers half the distance with speed of \$3 \ m/s\$. The other half of the distance is covered in two equal time interval with speed of \$4.5\ m/s\$ and \$7.5\ m/s\$ respectively. The average speed of the particle during this motion is

\$\displaystyle t_{1}=\dfrac{\dfrac{S}{2}}{3}=\dfrac{S}{6}\$\$\displaystyle \dfrac{S}{2}=4.5\times t_{2}+7.5t_{3}\Rightarrow 12t_{2}=\dfrac{S}{2}\$We have \$t_2=t_3\$Total time \$\displaystyle t=t_{1}+t_{2}+t_3=\dfrac{S}{6}+\dfrac{S}{24}+\dfrac{S}{24}=\dfrac{S}{4}\$Average of \$\displaystyle \dfrac{S}{t}=\dfrac{S}{\dfrac{S}{4}}=4\: m/s\$

Q: The position of an object moving along x-axis is given by x = a + \$bt^2\$, where a = 8.5 m and b = 2.5 m \$s^{-2}\$ and t is measured in seconds. The average velocity of the object between t = 2 s and t = 4 s is

Position is given as \$x = a+bt^2 = 8.5+2.5t^2\$Position at \$t = 2 \ s\$, \$x_2 = 8.5+2.5(2)^2 = 18.5 \ m\$Position at \$t = 4 \ s\$, \$x_1 = 8.5+2.5(4)^2 = 48.5 \ m\$Displacement \$S = x_2-x_1 = 48.5-18.5 = 30 \ m\$Time taken \$t = 4-2 = 2 \ s\$Average velocity \$V_{avg} = \dfrac{S}{t} = \dfrac{30}{2} = 15 \ m/s\$

Q: Joseph Jogs from one end \$A\$ to the other end \$B\$ of a straight \$300\ m\$ road in \$2\$ minutes \$30\$ seconds and then turns around and jogs \$100\ m\$ back to point \$C\$ in another \$1\$ minute. What are Joseph's average speeds and velocities in jogging (a) from \$A\$ to \$B\$ and (b) from \$A\$ to \$C\$?

From pont A to BDistance covered\$=300m\$Displacement\$=300m\$Time taken\$=2\ min\ 30\ sec=(2\times60)+30=150\ sec\$Average speed\$=\frac{Distance\ covered}{Time taken}=\frac{300}{150}=2\ m/sec\$Average velocity\$=\frac{Displacement}{Time taken}=\frac{300}{150}=2\ m/sec\$From point A to CDistance covered\$=300+100=400m\$Displacement\$=300-100=200m\$Time taken\$=3\ min\ 30\ sec=(3\times60)+30=210\ sec\$Average speed\$=\frac{Distance\ covered}{Time taken}=\frac{400}{210}=1.90\ m/sec\$Average velocity\$=\frac{Displacement}{Time taken}=\frac{200}{210}=0.95\ m/sec\$

Q: A car travels from A to B at a, speed of \$20\, km\, h^{-1}\$ and returns at a speed of \$30\, km\, h^{-1}\$. The average speed of the car for the whole journey is:

Let distance from A to B be xTime for travel from A to B= x/20Time for travel from B to A= x/30Hence,speed for total distance 2x is: \$\dfrac{2x}{\dfrac{x}{30}+\dfrac{x}{20}}=\dfrac{2x\times60}{5x}\$=24 km\hr

Q: A driver of a car travelling at \$52\$km \$h^{-1}\$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in \$5\$s. Another driver going at \$3\$km \$h^{-1}\$ in another car applies his brakes slowly and stops in \$10\$s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

(A) : Initial speed of the car A \$u = 52\$ \$kmh^{-1} = 52 \times \dfrac{5}{18} = 14.44\$ \$ms^{-1}\$ The car stops in 5 seconds i.e \$v = 0\$ at \$t = 5\$(A) : Initial speed of the car A \$u = 3\$ \$kmh^{-1} = 3 \times \dfrac{5}{18} = 0.83\$ \$ms^{-1}\$ The car stops in 10 seconds i.e \$v = 0\$ at \$t = 10\$With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.Area under the speed-time graph gives the distance covered.\$\therefore\$ Distance covered by car A \$x_A = \dfrac{1}{2}\times 14.44 \times 5 = 36.1 \$ m\$\therefore\$ Distance covered by car A \$x_B = \dfrac{1}{2}\times 0.83 \times 10 = 4.15 \$ mThus car A travels more distance than B.

Q: Abdul, while driving to school, computes the average speed for his trip to be \$20\$km \$h^{-1}\$. On his return trip along the same route, there is less traffic and the average speed is \$30\$km \$h^{-1}\$. What is the average speed for Abdul's trip?

Let us assume that Abdul is travelling \$x\$ kilometer distance while going to school.Distance = Speed x Time1)When it's average speed is 20 km/hr\${x}={v}_{1}\times{t}_{1}={20}\times{t}_{1}\$2)When it's average speed is 30 km/hr\${x}={v}_{2}\times{t}_{2}={30}\times{t}_{2}\$\${ V }_{ average }=\cfrac { total\quad distance\quad{travelled} }{ total\quad time\quad taken } \$\${ V }_{ average }=\cfrac { x+x }{ \frac { x }{ { v }_{ 1 } } +\frac { x }{ { v }_{ 2 } } } =\cfrac { 2x }{ \frac { x }{ 20 } +\frac { x }{ 30 } } =24\ {km\ h}^{-1}\$

Q: A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of the distance at speed v. If the average speed is 48 km/h, then the value of v is-

Given, \$Distance=200m,Speed =40km/h \$ Average speed =\$48km/h\$\$48=\dfrac{200}{\dfrac{100}{40}+\dfrac{100}{v}}\Rightarrow \dfrac{1}{40}+\dfrac{1}{v}=\dfrac1{24}\Rightarrow v=\dfrac{120}{5-3}=\dfrac{120}{2}\Rightarrow=60km/h\$

Q: A car travels half of the distance with constant velocity 40 km/h and another half with a constant velocity of 60 km/h along a straight line. The average velocity of the car is

Suppose total distance is \$2x\$ \$km\$ thentime taken in traveling the distance first \$x\$ will be \$t_1=\dfrac{x}{40}\$ and the time taken in traveling the distance second \$x\$ will be \$t_2=\dfrac{x}{60}\$ now the average velocity can be calculated as \$V=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{2x}{\dfrac{x}{40}+\dfrac{x}{60}}=\dfrac{2\times 40\times 60}{40+60}=48Km/h\$

Physics

definition

Velocity

Velocity is defined as the rate of change of displacement with time.
It is a vector quantity and has units m/s.
The direction of a velocity vector is same as the direction of the motion of body at that instant of time. Magnitude of instantaneous velocity is same as the speed at the instant.

definition

Average velocity

Average Velocity = \frac{Total Displacement}{Total time}

It is a vector quantity and has units of m/s.
Magnitude of average velocity is always less than or equal to the average speed because displacement is always smaller than or equal to distance.

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A car travels half of the distance with constant velocity 40 km/h and another half with a constant velocity of 60 km/h along a straight line. The average velocity of the car is

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Velocity

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A car moving on a straight road covers one third of a certain distance with 20 Km/h and the rest with 60 Km/h. The average speed is :

A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time interval with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is

The position of an object moving along x-axis is given by x = a + bt2, where a = 8.5 m and b = 2.5 m s−2 and t is measured in seconds. The average velocity of the object between t = 2 s and t = 4 s is

Joseph Jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?

A car travels from A to B at a, speed of 20kmh−1 and returns at a speed of 30kmh−1. The average speed of the car for the whole journey is:

Abdul, while driving to school, computes the average speed for his trip to be 20km h−1. On his return trip along the same route, there is less traffic and the average speed is 30km h−1. What is the average speed for Abdul's trip?

A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of the distance at speed v. If the average speed is 48 km/h, then the value of v is-

A car travels half of the distance with constant velocity 40 km/h and another half with a constant velocity of 60 km/h along a straight line. The average velocity of the car is

A particle moving in a straight line covers half the distance with speed of $3 m / s$ . The other half of the distance is covered in two equal time interval with speed of $4.5 m / s$ and $7.5 m / s$ respectively. The average speed of the particle during this motion is

The position of an object moving along x-axis is given by x = a + $b t^{2}$ , where a = 8.5 m and b = 2.5 m $s^{- 2}$ and t is measured in seconds. The average velocity of the object between t = 2 s and t = 4 s is

Joseph Jogs from one end $A$ to the other end $B$ of a straight $300 m$ road in $2$ minutes $30$ seconds and then turns around and jogs $100 m$ back to point $C$ in another $1$ minute. What are Joseph's average speeds and velocities in jogging (a) from $A$ to $B$ and (b) from $A$ to $C$ ?

A car travels from A to B at a, speed of $20 k m h^{- 1}$ and returns at a speed of $30 k m h^{- 1}$ . The average speed of the car for the whole journey is:

Abdul, while driving to school, computes the average speed for his trip to be $20$ km $h^{- 1}$ . On his return trip along the same route, there is less traffic and the average speed is $30$ km $h^{- 1}$ . What is the average speed for Abdul's trip?