Kinematics

Step 1: Use the equation of motion.
The equation for the distance traveled under uniform acceleration is: Since the body starts from rest, , so the equation becomes:
Step 2: Find the acceleration.
For the first 10 m in 10 seconds, we use:
Step 3: Calculate the distance after 30 seconds.
Now, using , calculate the distance traveled in 30 seconds:
Step 4: Conclusion.
Thus, the total distance traveled after 30 seconds is 90 meters, which corresponds to option (D).

Step 1: Understand the meaning of area under a velocity-time graph.
In a velocity-time graph, the area under the graph represents the displacement. This is because displacement is the integral of velocity over time.
Step 2: Conclusion.
Thus, the area under the velocity-time graph gives the displacement, corresponding to option (A).

Step 1: Use the formula for relative velocity.
When two velocities are perpendicular (normal to each other), the magnitude of the relative velocity of A with respect to B is given by:
Step 2: Conclusion.
Thus, the relative velocity of A with respect to B is , corresponding to option (C).

Step 1: Understanding the motion.
When a body is thrown vertically upward, the displacement equation for the height is given by: where is the initial velocity, is the acceleration due to gravity, and is the time. We need to find the height difference when the body passes through the same point at times and .
Step 2: Using the equation of motion.
At time , the height is: At time , the height is: The total height covered from the initial position to the point at times and is the sum of the two heights: Thus, the correct formula for the height at that point is .
Step 3: Conclusion.
The correct answer is (4) g , which gives the height when the body passes through the same point at two different times.

Step 1: Use the equations of motion.
For an object dropped from rest, the velocity at the point of impact is given by the equation: where is the acceleration due to gravity and is the height from which the body is dropped.
Step 2: Conclusion.
Thus, the velocity with which the body touches the ground is , corresponding to option (B).

Step 1: Use the equation of motion for the final velocity.
The final velocity of the stone can be calculated using the equation: where is the initial velocity, is the acceleration due to gravity, and is the height of the building.
Step 2: Calculate the final velocity.
Substitute the values:
Step 3: Conclusion.
Thus, the velocity with which the stone will reach the ground is , which corresponds to option (D).

Step 1: Use the equation of motion.
We know that the distance covered in uniformly accelerated motion is given by: where is the initial velocity, is the acceleration, and is the time.
Step 2: First 3 seconds.
For the first 10 m covered in 3 seconds:
Step 3: Next 3 seconds.
For the next 100 m covered in 3 seconds: Using the initial velocity from the previous part: Solving for gives us the acceleration. After calculation, we find the acceleration to be 9 m/s².

Step 4: Conclusion.
Thus, the acceleration is 9 m/s², corresponding to option (B).

Step 1: Formula for average speed.
The average speed is given by:
Step 2: Calculate total distance.
Total distance = m.
Step 3: Calculate total time.
Time taken to cover the first 10 m at 10 m/s: Time taken to cover the next 10 m at 20 m/s: Total time = .
Step 4: Calculate average speed.

Step 5: Conclusion.
Thus, the average speed is m/s, which corresponds to option (C).