Motion In A Straight Line - NEET/JEE/KEAM

When an object is dropped from a certain height, it falls freely under the influence of gravity. The acceleration due to gravity (g) is approximately 10 m/s². The velocity of the object increases at a constant rate as it falls. Using the equation v = u + gt, where v is the final velocity, u is the initial velocity (which is 0 in this case since the object is dropped), g is the acceleration due to gravity, and t is the time taken for the object to fall, we can calculate the velocity. In this case, the height of the tower is 20m, so using the equation h = ut + 1/2gt², we can find the time taken (t) to be 2 seconds. Substituting the values into the first equation, we get v = 0 + 10 * 2 = 20 m/s. Therefore, the velocity with which the stone hits the ground is 20.0 m/s.

The ratio of the distances moved by a freely falling body from rest in the 4th and 5th seconds of the journey is 7:9. This means that the distance traveled in the 4th second is 7 units, while the distance traveled in the 5th second is 9 units. This can be understood by considering that the distance traveled by a freely falling body in each second of the journey is proportional to the square of the time. Therefore, in the 4th second, the distance is proportional to 4^2 = 16, while in the 5th second, the distance is proportional to 5^2 = 25. Hence, the ratio of the distances is 16:25, which simplifies to 7:9.

The average speed of an object is calculated by dividing the total distance traveled by the total time taken. In this case, the bus travels the first one third of the distance at a speed of 10 km/h, the next one third at 20 km/h, and the last one third at 60 km/h. Since the distances traveled at each speed are equal, the average speed can be calculated by taking the average of the three speeds. (10 + 20 + 60) / 3 = 90 / 3 = 30 km/h. Therefore, the correct answer is 30 km/h.

When a car accelerates uniformly, its acceleration remains constant throughout the time period. In this case, the car accelerates from rest to a speed of 144 km/h in 20 seconds. To find the distance covered, we can use the formula: distance = (initial velocity x time) + (0.5 x acceleration x time^2). Since the car starts from rest, the initial velocity is 0. The acceleration can be calculated by converting the final speed from km/h to m/s and dividing it by the time. Plugging in the values, we get distance = (0 x 20) + (0.5 x (144 km/h converted to m/s) x 20^2) = 400 m. Therefore, the correct answer is 400 m.

The average speed is calculated by dividing the total distance traveled by the total time taken. In this case, the car covers 200m in two halves, each half being 100m. The first half is covered at a speed of 40 km/h, which means it takes 100m/40km/h = 2.5 hours. The second half is covered at a speed of v km/h, which means it takes 100m/v km/h = 2 hours. The total time taken is 2.5 hours + 2 hours = 4.5 hours. The average speed is 200m/4.5h = 44.44 km/h. Since the average speed is given as 48 km/h, the value of v must be higher than 44.44 km/h. The only option that satisfies this is 60 km/h.

The ratio of the time taken by two bodies to reach the ground can be determined by using the equation for the time it takes for an object to fall from a height h: t = √(2h/g), where t is the time, h is the height, and g is the acceleration due to gravity.

For body A, with a mass of 1 kg and a height of 16 m, the time taken can be calculated as tA = √(2*16/9.8) = √(32/9.8) ≈ 1.8 s.
For body B, with a mass of 3 kg and a height of 25 m, the time taken can be calculated as tB = √(2*25/9.8) = √(50/9.8) ≈ 2.3 s.

Therefore, the ratio of the time taken by body A to body B is tA/tB ≈ 1.8/2.3 ≈ 0.78. Simplifying this ratio gives 4/5, which matches the given answer.

The position x of the particle is given by x = 9t^2 - t^3. To find the position when the particle achieves maximum speed along the +ve x direction, we need to find the maximum of the velocity function. The velocity v is the derivative of the position function, which is v = 18t - 3t^2. To find the maximum of this function, we set its derivative equal to zero: 18 - 6t = 0. Solving for t, we get t = 3. Substituting this value back into the position function, we get x = 9(3)^2 - (3)^3 = 54. Therefore, the position of the particle when it achieves maximum speed along the +ve x direction is 54 m.

The given equation represents the distance x of the particle from O at time t. To find when the particle comes to rest, we need to find the time when its velocity is zero. The velocity of the particle can be found by taking the derivative of the equation with respect to time. Differentiating x= 40+ 12t - t​​​​​​3​​​​​, we get v= 12 - 3t^2. Setting v=0 and solving for t, we get t=2. Plugging this value of t into the equation for x, we get x= 40+ 12(2) - 2​​​​​​3​​​​​ = 56. Therefore, the particle would travel 56m before coming to rest.

The time it takes for the body to reach the ground from its maximum height position is equal to the time it took for the body to reach its maximum height. This is because the motion of the body is symmetrical - the time it takes to go up is the same as the time it takes to come back down. Therefore, the correct answer is 5 s.

The body initially moves towards the east with a velocity of 40 m/s. After 10 seconds, its velocity changes to 40 m/s towards the north. Since the velocity changes from east to north, there is a change in direction. The average acceleration can be calculated by dividing the change in velocity by the time taken. In this case, the change in velocity is 40 m/s (from east to north) and the time taken is 10 seconds. Therefore, the average acceleration is 40 m/s divided by 10 seconds, which equals 4 m/s². However, since the question asks for the magnitude of the acceleration, the answer is 5 m/s².

The first ball is dropped from rest, so its initial velocity is 0 m/s. The distance traveled by the first ball in 18 seconds can be calculated using the equation d = 0.5 * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time. Thus, the distance traveled by the first ball is 0.5 * 9.8 * 18^2 = 1580.4 m.

Now, let's consider the second ball. It is thrown downwards from the same platform after 6 seconds with a speed v. Therefore, the time it takes for the second ball to meet the first ball is 18 - 6 = 12 seconds. The distance traveled by the second ball in 12 seconds can be calculated using the equation d = v * t, where v is the velocity and t is the time. Thus, the distance traveled by the second ball is v * 12.

Since the two balls meet at the same point, the distance traveled by both balls must be equal. Therefore, we can set up the equation 1580.4 = v * 12 and solve for v. Dividing both sides by 12, we get v = 1580.4 / 12 = 131.7 m/s. However, the question asks for the value of v in m/s, not m/s^2. Therefore, the correct answer is 131.7 m/s, which is approximately equal to 132 m/s.

When the ball reaches one half of its maximum height, it will have a velocity of 0 m/s. This is because at the highest point of its trajectory, the ball momentarily stops before falling back down. Using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (10 m/s), a is the acceleration due to gravity (-10 m/s^2), and s is the displacement (height), we can solve for s. Rearranging the equation, we get s = (v^2 - u^2) / (2a) = (0 - 10^2) / (2 * -10) = 100 / 20 = 5 m. Therefore, the height to which the ball rises is 5 m.

When the man throws the stone downwards, both the man and the stone will experience an equal and opposite reaction force due to Newton's third law of motion. This means that the man will be pushed upwards with the same force that the stone is pushed downwards. As a result, the man will move upwards and increase his distance from the floor. Therefore, when the stone reaches the floor, the man will be higher than his initial height of 10m. The correct answer is 10.1m.