Question 51:
easy
Velocity versus time graph for a body projected vertically upwards is :-
Question 52:
moderate
A particle is thrown with the speed u at an angle α with the horizontal. When the particle makes an angle β with the horizontal, its speed will be :
During Projectile motion horizontal component of velocity remains same so,
\[ u cos\alpha = v cos \beta \]
\[ \frac{u cos\alpha}{cos \beta} = v \]
\[ {u .cos\alpha}.{sec \beta} = v \]
Question 53:
easy
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range?
Question 54:
moderate
If a body A of mass M is thrown with velocity v at an angle 30° to the horizontal and another body B of same mass is thrown with same speed at an angle of 60° to the horizontal, the ratio of range of A and B will be :
For Complementary angles ranges are equal
Question 55:
moderate
Which of the following statements is incorrect :
The statement is wrong because:
- Displacement is the shortest straight-line distance between an object's initial and final positions, and it has both magnitude and direction.
- Path length (or distance) is the total length of the path traveled by the object, regardless of direction.
Example:
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
Question 56:
moderate
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 ?
Question 57:
moderate
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**.
Question 58:
easy
The numerical ratio of distance to magnitude of displacement is :
The numerical ratio of distance to the magnitude of displacement depends on the type of motion:
1. For straight-line motion in one direction, the distance and displacement are the same, so the ratio is:
\[
\frac{\text{Distance}}{\text{Displacement}} = 1
\]
2. For any other type of motion (like a curved path or circular motion), the distance is generally greater than or equal to the displacement, making the ratio:
\[
\frac{\text{Distance}}{\text{Displacement}} \geq 1
\]
The ratio is greater than 1 because distance is the total path travelled, while displacement is the shortest straight line between the start and end points.
Question 59:
moderate
A car is moving along a straight line OP as shown in the figure. It moves from O to P in 18 s and returns from P to Q in 6 s. Which of the following statements is not correct regarding the motion of the car :
Total Distance = OP + PQ = 360 + 120 = 480 m
Displacement = OQ= 240 m
Average Speed = 480/24 = 20 m/s
Average Velocity= 240/24= 10 m/s
Question 60:
easy
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 s. Its average velocity is :