## Wednesday, October 26, 2016

### Motion in One Dimension Problems with Solutions Thirteen

We are solving series of problems in one dimensional motion. It is given in the problem that a point moves in a straightline under a retardation as given in the problem where k is a constant.Acceleration is defined as rate of change of velocity. If a body has acceleration means, its velocity is increasing with time. If it is deceasing with  time, it is called retardation.

It is given that the particle has initial velocity. We need to know the distance covered by it in the given time. The problem is as shown in the diagram below.

Solution

We know that retardation is a negative acceleration. Here in this case the velocity of the body is decreasing with time. To indicate it, a negative sign is taken into consideration. We can rearrange the velocity  in terms of time as shown in the diagram below. As we know the part of velocity, to get the total velocity by integrating it in the given limits. By using rules of integration, we can the value of final velocity after applying the limits of lower and upper.

We need displacement and hence write the velocity as the rate of change of displacement. We can write the equation for small displacement and by integrating it, we can get the displacement of the particle as shown in the diagram below.

Problem

In the arrangement shown in the figure and instantaneous velocities of two masses, we need to know the angle between the vertical as shown in the diagram below.

Solution

Let the vertical distance between pulley and the surface is X. Let the horizontal distance is 2a. It is given that the length of the string is constant and hence it can be found in the given terms as shown in the diagram below. As the equation is constant, its differentiation with time is zero.

By applying that condition, we can the relation between two velocities as shown. Basing on that we can find the angle and the detailed solutions is given below.

Problem

It is given in the problem that a ball is thrown vertically upward with a known velocity. While going upward and again while coming back after reaching its maximum height, the ball came back to the same point and the time interval between them is given to us in the problem. We need to measure the initial velocity of the ball. The problem is as shown in the diagram below.

Solution

Let the height that is crossed twice is and the intervals are given in the problem as shown in the diagram. Thus we can equate the height equation both in terms of the first time and the second time. By solving these two  equations, we can solve the problem as shown in the diagram below.

Problem

It is given in the problem that a body is projected vertically and different points are given in the journey. If the body is released from position A and we need to know the time of descents ratio need to be found.

Solution

We know that the time intervals for each part is same and hence to reach each point, the time taken in multiples of that interval as shown in the diagram below. We need to use the second equation of motion and solve the problem as shown in the diagram below.

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