13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Solution:
The distance-time graph can be plotted as follows.
When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is at rest.
14. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Solution:
The speed-time graph can be plotted as follows.
Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.
15. What is the quantity which is measured by the area occupied below the velocity-time graph?
Solution:
Considering an object in uniform motion, its velocity-time graph can be represented as follows.
Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA x OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:
Area under the velocity-time graph = velocity x time.
Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.
16. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
(a) Given, the bus starts from rest. Therefore, initial velocity (u) = 0 m/s
Acceleration (a) = 0.1 m.s-2
Time = 2 minutes = 120 s
Acceleration is given by the equation a=(v-u)/t
Therefore, terminal velocity (v) = (at)+u
= (0.1 m.s-2 x 120 s) + 0 m.s-1
= 12 m.s-1 + 0 m.s-1
Therefore, terminal velocity (v) = 12 m/s
(b) As per the third motion equation, 2as = v2 – u2
Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120 s, the following value for s (distance) can be obtained.
Distance, s = (v2 – u2)/2a
=(122 – 02)/2(0.1)
Therefore, s = 720 m.
The speed acquired is 12 m.s-1 and the total distance travelled is 720 m.
17. A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of –0.5 m s-2. Find how far the train will go before it is brought to rest.
Solution:
Given, initial velocity (u) = 90 km/hour = 25 m.s-1
Terminal velocity (v) = 0 m.s-1
Acceleration (a) = -0.5 m.s-2
As per the third motion equation, v2 - u2 = 2as
Therefore, distance traveled by the train (s) = (v2 - u2)/2a
s = (02 - 252)/2(-0.5) meters = 625 meters
The train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.
18. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
Solution:
Given, initial velocity (u) = 0 (the trolley begins from the rest position)
Acceleration (a) = 0.02 ms-2
Time (t) = 3s
As per the first motion equation, v = u + at
Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms-2)(3s) = 0.06 ms-1
Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s-1
19. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10s after start?
Solution:
Given, the car is initially at rest; initial velocity (u) = 0 ms-1
Acceleration (a) = 4 ms-2
Time period (t) = 10 s
As per the second motion equation, s = ut + 1/2 at2
Therefore, the total distance covered by the car (s) = 0 x 10m + 1/2 (4ms-2)(10s)2
= 200 meters
Therefore, the car will cover a distance of 200 meters after 10 seconds.
20. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
Given, initial velocity (u) = 5 m/s
Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)
Acceleration = 10 ms-2 in the direction opposite to the trajectory of the stone = -10 ms-2
As per the third motion equation, v2 – u2 = 2as
Therefore, the distance travelled by the stone (s) = (02 – 52)/ 2(10)
Distance (s) = 1.25 meters
As per the first motion equation, v = u + at
Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a
= (0-5)/-10 s
Time taken = 0.5 seconds
Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.
21. 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?
Solution:
Given, diameter of the track (d) = 200m
Therefore, the circumference of the track (π x 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 x 200π/40 meters
= (140 x 200 x 22)/(40 x 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.
22. 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?
Solution:
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