A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn/Sn+1 is :
Sn = u + a/2(2n-1) = a/2(2n-1)
Sn+1 = u + a/2(2(n+1)-1) = a/2(2n+1)
Sn / Sn+1 = (2n-1)/(2n+1)
A body moves 6 m north, 8 m east and 10 m vertically upwards, what is its resultant displacement from initial position?
1. Movement 1: 6 m North
\[
\text{Displacement}_1 = 6 \hat{j}
\]
2. Movement 2: 8 m East
\[
\text{Displacement}_2 = 8 \hat{i}
\]
3. Movement 3: 10 m Vertically Upwards
\[
\text{Displacement}_3 = 10 \hat{k}
\]
Total Displacement:
\[
\text{Total Displacement} = 6 \hat{j} + 8 \hat{i} + 10 \hat{k}
\]
Magnitude of Displacement:
\[
|\text{Total Displacement}| = \sqrt{(8^2) + (6^2) + (10^2)} = \sqrt{64 + 36 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m}
\]
A person moves 30 m north and then 20 m towards east and finally 30√2 min south -west direction. The displacement of the person from the origin will be
1. Movement 1: 30 m North
\[
\text{Displacement}_1 = 30 \hat{j}
\]
2. Movement 2: 20 m East
\[
\text{Displacement}_2 = 20 \hat{i}
\]
3. Movement 3: \(30\sqrt{2}\) m Southwest
\[
\text{Displacement}_3 = -30 \hat{i} - 30 \hat{j}
\]
Total Displacement:
\[
\text{Total Displacement} = (20 \hat{i} + 30 \hat{j}) + (-30 \hat{i} - 30 \hat{j}) = -10 \hat{i} + 0 \hat{j}
\]
Displacement = 10 m West
Which of the following statements is incorrect :
The statement is wrong because:
If an object moves in a circular path and returns to its starting point, the displacement is 0 (since the initial and final positions are the same), but the path length is the total distance traveled around the circle.
Thus, displacement is not always equal to the path length
The three initial and final positions of a man on the x-axis are given as :-
(i) (–3m, 7m)
(ii) (7m, –3m)
(iii) (–7 m, 3m)
Which pair gives the negative displacement ?
A drunkard is walking along a straight road. he takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after :
### Given:
- The drunkard takes **5 steps forward** (5 meters) and **3 steps backward** (3 meters).
- Each step is **1 meter** and takes **1 second**.
- There is a pit **11 meters** away.
### Correct Solution:
1. In **1 full cycle** (5 steps forward + 3 steps backward), the net distance covered is:
\[
5 - 3 = 2 \text{ meters}.
\]
This cycle takes **8 seconds**.
2. We need to find out how long it takes the drunkard to fall into the pit which is **11 meters** away.
### Step-by-step process:
- After **1 cycle** (8 seconds), the drunkard is at **2 meters**.
- After **2 cycles** (16 seconds), the drunkard is at **4 meters**.
- After **3 cycles** (24 seconds), the drunkard is at **6 meters**.
Now, in the **next (4th) cycle**, the drunkard will walk forward:
- After the first **5 steps forward** (which takes 5 seconds), he will move from **6 meters** to **11 meters**, falling into the pit.
### Total time:
\[
24 \text{ seconds} + 5 \text{ seconds} = 29 \text{ seconds}.
\]
Conclusion:
The drunkard will fall into the pit after **29 seconds**.
P is the point of contact of a wheel and the ground. The radius of the wheel is 1m. The wheel rolls on the ground without slipping. The displacement of point P when the wheel completes half rotation is :
Horizontal Displacement = πR
Vertical Displacement = 2R
Total Displacement = ((πR)^2+(2R)^2)^1/2=
\[\sqrt{\pi^{2}+4} m\]