Motion

Question: 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20s?

Solution:

Given, diameter, d of the circular track = 200 m.

Therefore, radius, r = 200 m/2 = 100 m

Time, t to complete one round = 40 s

Thus, in 2 minute 20 s = 2 × 60 + 20 = 140 s

Distance covered, s =?

Displacement =?

Circumference of the circular path is the distance covered by athletes to complete one round.

We know that, circumference of circle = `2pir`

∴ Circumference = 2xxpixx100\ m`

∴ Circumference = 200\ pi\ m`

Now, ∵ In 40 s athlete cover a distance of `200\ pi\ m`

∴ In 1 s distance covered by the athlete `=(200\ pi\ m)/(40\ s)`

∴ In 140 s distance covered by the athlete `= (200\ pi\ m)/(40\ s)xx140\ s`

`=(200\ m xx 140\ sxx22)/(40\ sxx7)`

`=5\ m xx 20 xx 22 = 2200\ m`

Now, number of rounds `= (text{Total distance})/(text{circumference})`

`=(2200\ m)/(200\ pi)`

`=(2200\ m)/(200 xx 22//7)`

`=(2200\ m xx7)/(200xx22)=3.5` round

Thus, in given time of 2 minute 20 second, athlete will complete 3.5 rounds to the circular trak.

3.5 rounds means three rounds and half round. This means after the 3.5 rounds athlete will be just opposite to the starting point.

Just opposite to the starting point means, the displacement is equal to the diameter of circular path.

Thus, distance covered = 2200 meter and displacement = 200 m Answer

Question: 2. 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?

Answer:

Here given, distance covered (From A to B), s = 300 m

Time taken to cover this distance, `t` = 2 minute 30 second

`=>t= 2xx60+30=150\ s`

Distance from A to C, `s` = 300 m + 100 m = 400 m.

Time taken to cover distance of 100 m (B to C) = 1 minute = 60 second

Thus, Total time taken to cover Distance from A to B and from B to C, i.e. from A to C

= 150 s + 60 s = 210 s

Thus, Average speed and velocities =?

Calculation of Average speed from A to B

We know that, Average speed, `v` = Total distance covered/Total Time taken

`=(300\ m)/(150\ s)`

`=>v = 2\ m//s`

Calculation of Average velocity from A to B

Since, the given path is straight, and distance is equal to displacement thus average velocity will be equal to the average speed.

Thus, displacement = 300 m and time taken = 150 s

We know that, Average velocity, `v_(av)` = Displacement / Total time taken

`v_(av) = (300\ m)/(150\ s)=2\ m//s`

Thus, here average speed = average velocity = 2 m/s

Calculation of Average speed from A to C

Distance covered from A to C = Distance of A to B + Distance of A to C

∴ Total distance covered = 300 m + 100 m = 400 m

Total time taken = time taken from A to B + Time taken from A to C

Time taken = 150 s + 60 s = 210 s

We know that average speed, `v` = Total distance covered/ Total time taken

`=>v = (400\ m)/(210\ s)`

`=>v = 1.9\ m//s`

Calculation of average velocity from A to C

Here displacement = distance from A to C = 300m – 100 m = 200 m

Total time taken = time taken from A to B + Time taken from A to C

Time taken = 150 s + 60 s = 210 s

Now, we know that Average velocity, `v_(av)` = Displacement/Total time taken

`=>v_(av) = (200\ m)/(210\ s)`

`=>v_(av) = 0.95\ m//s`

Thus, average speed and velocity from A to B is equal to 2 m /s

And Average speed from A to C = 1.9 m/s and average velocity from A to C = 0.95 m/s Answer

Question: 3. 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?

Answer:

Given, Average speed of Abdul trip while driving to the school = 20 km/h

And Average speed of Abdul while returning from school = 30 km/h

Thus, Overall average speed=?

Let, distance of the school = s km

Thus, total distance of going and returning from school = 2s km

And time taken to go to school = t₁ h

And time taken to return from school = t₂ h

As given, Average speed while going to the school = 20 km/h

We know that, Average speed = Total distance / time taken

`=>20\ km// h = (s\ km)/t_1`

`=>t_1= (s\ km)/ (20\ km//h)`

`=>t_1=s/20\ h`

Again as given Average speed while returning from school

We know that, Average speed = Total distance / time taken

`=>30\ km// h = (s\ km)/t_2`

`=>t_2= (s\ km)/ (30\ km//h)`

`=>t_2=s/30\ h`

Total time taken while going and returning from school

`=>t_1+t_2=s/20\h+s/30\h`

`=>t_1+t_2=(3\ s+2\ s)/60\ h`

`=>t_1+t_2=(5\ s)/60\h`

Now, Average speed = Total distance travelled/Total time taken

`=(2\ s\ km)/(t_1+t_2)`

`=(2s\ km)/((5\ s)//(60)h)`

`=(2s\ km xx60)/(5\ s)`

`=(120\ s\ km)/(5\ s)`

`=24\ km//h`

Thus, Average speed of Abdul's trip = 24 km /h. Answer.

Question: 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^–2 for 8.0 s. How far does the boat travel during this time?

Answer:

Given, Initial speed, `u=0` [As motorboat is starting from rest]

Acceleration, `a=3.0\ m\ s^(-2)`

Time, `t=8.0\ s`

Thus, distance, s = ?

We know that, `s = ut + 1/2 at^2`

`=>s = 0xx8\ s + 1/2xx 3\ m\ s^(-2)xx(8\ s)^2`

`=>s=1/2xx3\ m\ s^(-2)xx64\ s^2`

`=>s = 3\ m\ s^(-2)xx32\ s^2`

`=>s = 96 m`

Thus, distance cover in the given time = 96 m. Answer

Question: 5. 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 farter after the brakes were applied?

Solution:

Given,

For first car, time, `t_1` = 5 s

Velocity, `u_1 = 52\ km\ h(^-1)`

`=(52xx1000)/(60xx60)\ m\ s^(-1)`

`=>u_1=14.44\ m s^(-1)`

For second car,

Time, `t_2` = 10 s

Velcoity, `u_2 = 3\ km\ h^(-1)`

`=(3xx1000)/(60xx60)\ m\ s^(-1)`

`=>u_2=0.83\ m\ s^(-1)`

Since, both cars come to rest after applying of brake, thus, `v_1=0` and `v_2=0`

We know that area under velocity time graph gives the distance covered by object in motion.

Thus, Area of triangle OCA will give the distance covered by first car.

Thus, area of triangle OCA `=1/2 xxOAxxOC`

`=1/2xx14.44\ m\ s^(-1)xx5\ s`

`= 36.1 m`

Thus, distance covered by Ist car = 36.1 m

Now, Area of triangle ODB will give the distance covered by second car.

Thus, Area of triangle ODB `= 1/2 xx OB xx OD`

`=1/2xx0.83\ m\ s^(-1)xx10\ s`

`=0.83\ m\ s^(-1)xx5\ s`

`=4.15\ m`

Thus, distance covered by second car, which comes to rest after 10 s = 4.15 m

Thus, first car, which was travelling with 52 km/h will travel farther after applying brakes. Answer

Question: 6. Figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

Solution:

According to graph at distance axis 1 square = 4/7 km

And at time axis 1 square = 0.4/5 hour

(a) Which of the three is travelling the fastest?

Steep of slope in the graph show the speed of object when distance is considered along `y`-axis.

Object which has most steep slow is the fastest.

In the graph it is clear that, object B has most steep slope, thus object B is the fastest among given object.

Alternate method

For B distance = 20.5 square

`= 20.5 xx 4/7\ km = 11.71\ km`

Time = 18 square

`= 18 xx 0.4/5` hour

`= 1.44` hour

Now we know that speed = Distance/Time

`= (11.71\ km)/(1.44\ h)`

`= 8.13\ km//h`

For object A

Distance = 9.5 square

`= 9.5 xx 4/7\ km`

`= 5.42\ km`

And time = 25 square

`=25xx0.4/5` hour

`= 2\ h`

Now, Speed = Distance/Time

`=(5.42\ km)/(2\ h)`

`=2.71\ km//h`

For C

Distance = 17 square

`=17 xx 4/7\ km`

`= 68/7= 9.71\ km`

Time = 23 square

`=23xx(0.4)/5` hour

`=1.84` hour

Now, speed = Distance /Time

`= (9.71\ km)/(1.84\ h)`

`= 5.2\ km//h`

Thus, object B is the fastest.

(b) Are all three ever at the same point on the road?

No, none object are ever at the same point on the road as it is clear from graph.

(c) How far has C travelled when B passes A?

Object B passes A at point m

At this point according to graph, C is at point n

Thus, distance covered by C = about 10 square

`=10xx 4/7\ km`

`= 5.71 \ km`

Thus, C travelled 5.71 km when B passes A

(d) How far has B travelled by the time it passes C?

B passes C at point p

At this point distance = 9 square

`= 9 xx 4/7` km

`=5.14` km

Thus, B travelled 5.14 km when it passes C.

Question: 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^–2, with what velocity will it strike the ground? After what time will it strike the ground?

Solution:

Given, initial velocity, `u=0` [Since before drop ball was at rest]

Distance covered, `s` = 20 m

Acceleration, `a = 10\ m\ s^(-2)`

Thus, final velocity, `v` = ?

And time taken to reach the ground, `t` = ?

We know that, `v^2 = u^2 + 2as`

`=>v^2 = 0^2 + 2xx10\ m\ s^(-2)xx20\ m`

`=>v^2 = 400\ m^2\ s^(-1)`

`=>v = sqrt (400\ m^2\ s^(-1))`

`=>v = 20\ m//s`

Now, we know that, `v=u+at`

`=>20\ m//s = 0 + 10\ m\ s^(-2)xxt`

`=>10\ m\ s^(-2)xxt = 20\ m//s`

`=>t = (20\ m//s)/(10\ m//s)`

`=>t = 2\ s`

Thus, velocity of the ball when strike at the ground will be 20 m/s and time taken to strike the ground = 2 second Answer

Question: 8. The speed-time graph for a car is shown in figure.

(a) Find how far does the car travel in the first 4 seconds? Shade the area on the graph that represents the distance travelled by the car during the period.

Solution:

The area of graph gives the distance travelled by an object.

In the given, graph at the interval of 4 second there are about 56 full and 12 half squares.

Thus, total number of squares `=56+12/2=62` squares.

Thus, area of these 62 squares will give the magnitude of distance travelled by car.

On the time axis, 5 squares = 2s

∴ 1 square `=2/5\ s`

And on the speed axis, 3 squares = 2 m/s

∴ 1 square = 2/3 m/s

Now, area of 1 square `=2/5\sxx2/3\m//s = 4/15\m`

Thus, area of 62 squares `=4/15\ mxx62`

`=248/15\ m = 16.53\ m`

Thus, distance travelled by car in first 4 seconds = 16.53 m. Answer

(b) Which part of the graph represents uniform motion of the car?

Answer: The part PQ of the graph shows the uniform motion of the car, as this portion is a straight line.

Question: 9. State which of the following situations are possible and give an example for each of the these:

(a) an object with constant acceleration but with zero velocity.

Answer: Yes, an object with constant acceleration but with zero velocity is possible.

While a stone is thrown in upward direction, its velocity becomes zero after reaching a maximum height. At this moment, stone starts falling because of acceleration due to gravity.

Thus, here initial velocity of the stone is zero, but acceleration is due to gravitational pull.

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: While circular motion, the direction of the object is towards the tangent of the circle while the acceleration is towards radius of the circle and since radius and tangent are perpendicular to each other.

Thus, in circular motion object is moving in a certain direction with an acceleration in the perpendicular direction.

Same situation happens in the case of projectile motion, as projectile motion is also an example of circular motion.

Question: 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Solution:

Given, Radius of the circular path = 42250 km

And time taken to complete one rotation = 24 hours

Thus, speed =?

We know that, circumference `=2pir`

Thus, circumference of the given orbit `=2xxpixx42250` km

Now, we know that, speed = Distance/Time

Here, circumeferece of the orbit = distance of the path

Thus, speed `= (2xxpixx42250\ km)/(24\ h)`

`=(2xx22xx42250\ km)/(7xx24\ h)`

=11065.47 km/h

Thus, speed of the satellite is equal to 11065.47 km/h. Answer