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 20 s?
Here we have,
Diameter = 200 m, therefore, radius = 200m/2 = 100 m
Time of one rotation = 40s
Time after 2m20s = 2 x 60s + 20s = 140s
Distance after 140 s = ?
Displacement after 140s =?
(a) Distance after 140s
We know that,distance=velocity ×time
⇒distance=15.7m/s ×140 s = 2198 m
(b) Displacement after 2 m 20 s i.e. in 140 s
Since,rotatin in 40 s=1
Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m.
Distance covered in 2 m 20 s = 2198 m
And, displacement after 2 m 20 s = 200m
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?
Here we have,
Distance from point A to B = 300 m
Time taken = 2 minute 30 second = 2 x 60 + 30 s = 150 s
Distance from point B to C = 100 m
Time taken = 1 minute = 60 s
(a) Average speed and velocity from point A to B
(b) Average speed and velocity from B to C
Therefore,average velocity=1.66 m/s west
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Strategy: We need to calculate the time taken in each of the trip. After that, we can calculate the average speed.
Let the distance of the school = s km
Let time to reach the school in first trip = t1
Let time to reach the school in second trip = t2
Therefore, average speed of Adbul = 24 km/h
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time?
Here we have,
Initial velocity (u) = 0
Acceleration (a) = 3.0m/s2
Time = 8 s
Therefore, distance (s) covered =?
Therefore, boat travel a distance of 96 m in the given time.
5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h 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?
Given for first driver,
In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.
Distance is calculated by the area under the slope of the graph.
6. Fig 8.11 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?
(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.
(b) All of them never come at the same point at the same time.
(c) According to graph; each small division shows about 0.57 km.
A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km
Thus, at this point C travels about
9.14 – (0.57 x 3.75) km = 9.14 km – 2.1375 km = 7.0025 km ≈ 7 km
Thus, when A passes B, C travels about 7 km.
(d) B passes C at point Q at the distance axis which is ≈ 4km + 0.57km x 2.25 = 5.28 km
Therefore, B travelled about 5.28 km when passes to C.
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s2, with what velocity will it strike the ground? After what time will it strike the ground?
Here we have,
Acceleration,a= 10 m s-2
8. The speed-time graph for a car is shown is Fig. 8.12.
(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.
(b) Which part of the graph represents uniform motion of the car?
(a) Distance travelled by car in the 4 second
The area under the slope of the speed – time graph gives the distance travelled by an object.
In the given graph
56 full squares and 12 half squares come under the area slope for the time of 4 second.
Total number of squares = 56 + 12/2 = 62 squares
The total area of the squares will give the distance travelled by the car in 4 second.
(b) Part MN of the slope of the graph is straight line parallel to the time axis, thus this potion of graph represents uniform motion of car.
9. State which of the following situations are possible and give an example for each of these:
(a) An object with a constant acceleration but with zero velocity
(b) An object moving in a certain direction with an acceleration in the perpendicular direction.
(a) The term acceleration implies that the velocity of the object is changing; inspite of that constant acceleration with zero velocity is impossible. When an object is thrown in upward direction, at the maximum height the velocity of the object becomes zero but still in that condition a constant acceleration due to gravity is working.
(b) Object moving in a certain direction with an acceleration in perpendicular direction is possible; in case of circular motion. When an object moves on a circular path, its direction is along the tangent of the circle but acceleration is towards the radius of the circle. We know, that a tangent always makes a right angle with the radius; so when an object is in circular motion, the acceleration and velocity are in mutually perpendicular direction.
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.
Here we have,
Radius, r = 42250km
Time, t = 24 hours