Chapter 7: Motion

Given, diameter of the track (d) = 200m

Therefore, the circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of rounds completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms^-1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms^-1= 1.9 m/s

Average velocity while traveling from A to B =300/150 ms^-1= 2 m/s

Average velocity while traveling from A to C =200/210 ms^-1= 0.95 m/s

Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t₁

Time taken to reach home = t₂

therefore, average speed while going to school = total distance travelled/ total time taken = d/t₁ = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t₂= 30 kmph

Therefore, t₁ = d/20 and t₂ = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t₁+t₂)kmph = 2d/(d/20+d/30)kmph

= 2/[(3 + 2)/60]

= 120/5 kmh^-1 = 24 kmh^-1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms^-2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at²

Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8)²

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h^-1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms^-1) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h^-1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms^-1) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) travelled farther post the application of brakes.

(a) Since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When B passes A, the distance between the origin and C is 8km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms^-2

As per the third motion equation,

v² – u² = 2as

Therefore,

= 2*(10ms^-2)*(20m) + 0

v² = 400m²s^-2

Therefore, v= 20ms^-1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

Therefore, t = (v-u)/a

= (20-0)ms^-1 / 10ms^-2

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

(a)

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6^th to the 10^th second.

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km

Time is taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h^-1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.