Describing Motion Around Us
1. Distance and Displacement of the Father
Total Distance Travelled:
Home to shop = 250m
Shop to home (to get bag) = 250m
Home to shop (again) = 250m
Shop to home (final return) = 250m
TotalDistance=250+250+250+250=1000m(or1km).
Displacement from Home:
Since the father finally returns back to his home, his initial and final positions are identical.
Displacement=0m.
2. Vertical Distance and Displacement of a Student
Height of each floor = 3m.
(i) Total vertical distance travelled:
Ground floor to 4th floor = 43=12m.
4th floor down to 2nd floor = (4-2)3=6m.
Totalverticaldistance=12+6=18m.
(ii) Displacement from the starting point:
Initial position = Ground floor (0m).
Final position = 2nd floor (23=6m above ground).
Displacement=6m-0m=6m (upwards).
3. Acceleration with Constant Speedometer Reading
Yes, it is possible.
A speedometer measures the magnitude of velocity (speed). Acceleration is defined as the rate of change of velocity (which is a vector quantity having both magnitude and direction). If the scooter is moving along a curved path or turning a corner at a constant speed, its direction of motion is continuously changing, which results in a change in velocity and thus produces acceleration (centripetal acceleration).
4. Average Acceleration and Distance Travelled by a Car
Given: Initial velocity u=0ms-1, Final velocity v=24ms-1, Time t=6s.
Average Acceleration (a):
a=v-ut=24-06=4ms-2
Distance Travelled (s):
s=ut+12at2=0(6)+12(4)(6)2=72m
5. Acceleration and Time Taken to Stop a Motorbike
Given: Initial velocity u=28ms-1, Final velocity v=0ms-1 (stops), Distance s=98m.
Acceleration (a): Using the kinematic equation v2=u2+2as:
02=282+2a98
0=784+196a
196a=-784
a=-4ms-2
(The acceleration/retardation is 4ms-2).
Time Taken (t): Using v=u+at:
0=28+(-4)t
4t=28
t=7s
6. Position-Time Graph Analysis (Fig. 4.27)
Do objects A and B ever have equal velocity? No.
Justification: In a position-time graph, the slope of the line represents velocity. Line A is steeper than line B throughout the graph, meaning object A’s velocity is constantly greater than object B’s velocity. They only cross each other at t=5s, meaning they are at the same position at that instant, but their velocities are not equal.
7. Position-Time Graph Analysis (Fig. 4.28)
Correct Options: (i) and (iii).
- (i) The average velocity of both over the 10s interval is equal because they have the same initial and final positions in the same time interval.
- (iii) The average speed of A is lower than that of B because object A travels in a straight line (shortest path), whereas object B follows a longer, winding path in the same duration, resulting in a greater total distance covered.
8. Distance Travelled by a Truck
Given: Initial speed u=54kmh-1=15ms-1, Final speed v=36kmh-1=10ms-1, Time t=36s.
Distance Travelled (s):
s=u+v2t=15+10236=12.536=450m
9. Total Distance Travelled by a Car Across Three Phases
Phase 1 (Acceleration): u=0, v=20ms-1, t1=5s
s1=0+2025=50m
Phase 2 (Constant Speed): v=20ms-1, t2=10s
s2=2010=200m
Phase 3 (Braking to a stop): u=20ms-1, v=0, t3=6s
s3=20+026=60m
Total Distance: stotal=50+200+60=310m.
10. Bus Stopping Distance and Obstacle Safety Check
Given: Initial speed u=36kmh-1=10ms-1, Reaction time tr=0.5s, Deceleration a=-2.5ms-2, Obstacle distance = 30m.
Distance during reaction time (s1):
s1=utr=100.5=5m
Distance during braking (s2): Using v2=u2+2as2:
02=102+2(-2.5)s2
0=100-5s2
5s2=100
s2=20m
Total Stopping Distance: 5m+20m=25m.
Conclusion: Yes, the bus will be able to stop safely 5m before reaching the obstacle (25m<30m).
11. Is an Object Kept on Earth at Rest?
Discussion: Rest and motion are relative terms. An object kept on the Earth (such as a desk or chair) is considered at rest relative to the Earth because its position does not change with respect to its surroundings on the planet. However, because the Earth itself rotates on its axis and revolves around the Sun, the object is simultaneously in motion relative to the Sun, stars, and space. Thus, it is at rest relative to terrestrial frames of reference, but in motion relative to celestial ones.
12. Cyclist Velocity-Time Graph (Fig. 4.30)
Shading Areas:
- (i) Constant velocity region: From t=20s to t=100s (rectangle of height 3ms-1).
- (ii) Decreasing velocity region: From t=100s to t=120s (trapezium/triangle dropping from 3 to 2ms-1).
Total Displacement (Total Area under Graph):
- Region 1 (0 to 20s): 12203=30m
- Region 2 (20 to 100s): 803=240m
- Region 3 (100 to 120s): 12(3+2)20=50m
- TotalDisplacement=30+240+50=320m.
Average Acceleration (over 120s):
aavg=v-ut=2-0120=160ms-20.0167ms-2
13. Marathon Runner Distance Estimation (Fig. 4.31)
By calculating the total area under the velocity-time graph (dividing the curve into manageable segments of constant/average speeds over time intervals up to 6.5hours):
Segment 1 (0 to 1.5h at 7.25kmh-1) 10.88km
Segment 2 (1.5 to 3h at 7.5kmh-1) =11.25km
Segment 3 (3 to 5.5h dropping from 7.5 to 6.5kmh-1) 17.5km
Segment 4 (5.5 to 6.5h at 6.5kmh-1) =6.5km
Total Estimated Distance 46.1km (consistent with a full marathon length plus warm-up/cool-down).
14. Car Displacement on State Highway
Phase 1 (Constant Velocity): v=6ms-1 for t1=120s (2min)
s1=6120=720m
Phase 2 (Acceleration): u=6ms-1, a=1ms-2 for t2=6s
s2=(66)+12(1)(6)2=36+18=54m
Total Displacement: 720+54=774m.
15. Motion of Two Cars (A and B)
Car A:
- Initial velocity (u) = 0ms-1, Final velocity (v) = 5ms-1, Time (t) = 5s
aA=5-05
aA=1ms-2
Meaning: Car A accelerates from rest to 5ms-1 in 5seconds at a rate of 1ms-2.
- Displacement in 5s = 12baseheight=1255=12.5m.
Car B:
- Initial velocity (u) = 0ms-1, Final velocity (v) = 3ms-1, Time (t) = 10s
aB=3-010
aB=0.3ms-2
- Displacement in 10s = 12baseheight=12103=15m.
16. Tip of a Clock’s Minute Hand (6:00 PM to 7:30 PM)
Given Time Interval: 1.5hours=90minutes=5400seconds.
Number of Rotations: 1.5 complete rotations (540).
Radius (r): 7cm=0.07m.
(i) Distance travelled:
Distance=1.5(2r)=1.522277=66cm=0.66m
(ii) Displacement: After 1.5 rotations, the tip is positioned diametrically opposite to its starting point. Thus, displacement equals the diameter:
Displacement=2r=14cm=0.14m
(iii) Speed:
Speed=DistanceTime=66cm5400s0.0122cms-1
(iv) Velocity:
Velocity=DisplacementTime=14cm5400s0.00259cms-1
(directed along the line connecting the initial and final positions).
