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>>Class 11
>>Physics
>>Motion in a Straight Line
>>Speed and Velocity
>> $Instantaneous speed, velocity, and accel...$

Instantaneous speed, velocity, and acceleration

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 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\$

Q: The displacement time graph for two particles A and B are straight lines inclined at angles of \$30^o\$ and \$60^0\$ with the time axis. The ratio of velocities of \$V_A:V_B\$ is

As velocity \$v=\dfrac{dx}{dt}=slope\$ of \$x-t\$ graph.so \$V_A:V_B=slope_A:slope_B=tan 30^0:tan 60^0=\dfrac{1}{\sqrt3}:\sqrt3=1:3\$

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