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\\)

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\\)

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\\)

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\\)

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

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.

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}\\)

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\\)

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\\)

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\\)

Velocity - Time Graph | Definition, Examples, Diagrams

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Average velocity from velocity-time graph

Instantaneous velocity at a given time-instant for a particle can be found from the y-intercept of its velocity-time graph.
From a particle's velocity-time graph, its average velocity can be found by calculating the total area under the graph and then dividing it by the corresponding time-interval.
For a particle with uniform acceleration, velocity-time graph is a straight line. Its average velocity is given by

v_{a v g} = (v_{i} + v_{f}) / 2

. From the graph, this can be found by drawing the y-intercepts of initial and final velocities and then drawing the mid-point.
In the given graph, instantaneous velocity

= v_{1}

t = t_{1}

and

v_{2}

t = t_{2}

.
Average velocity is found at E, i.e. mid-point of C and D.

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Acceleration from velocity-time graph

Acceleration of a particle is equal to the slope of a velocity-time graph.

a = \frac{v _{2} - v _{1}}{t _{2} - t _{1}}

In the given graph,

a = \frac{4 0 - 2 0}{4 - 2} = 10 m / s^{2}

Acceleration may be positive, negative or zero as it is a vector quantity.

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Displacement from velocity-time graph

Displacement of a particle in a given time-interval is equal to the total area under the velocity-time graph in the given time-interval.
In the given graph, displacement is found as area of trapezium ABCD.

s = \frac{1}{2} \times D C \times (A D + B C)

Note: Since, displacement is a vector quantity, Area below the time-axis is considered as negative and above it is considered as positive.

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Area under acceleration time graph

Change in velocity of a particle can be evaluated as area under the acceleration-time graph.

as

a = \frac{d v}{d t}

d v = a d t

V e l o c i t y = \int a d t

(Which is area under the curve.)

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Velocity Time graph for rest, uniform motion and uniform acceleration

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