Saturday, October 15, 2016

Motion in One and Dimension Problems with Solutions Four

We would like to start solving a problem related to horizontal projection. Here in this case we can project a body from a certain height with a initial velocity horizontally so that it can travel to a certain distance and strikes the ground. In this case also the body takes projectile path. The problem says about two cases where the heights,mass of the stones are different and the horizontal distance where they strikes the ground is different. We would like to find the relation between their velocity of projection. The problem is as shown in the diagram below.


To study the vertical motion of the body, we can use similar second equation of motion like one dimensional body as the horizontally projected body has no initial velocity in the vertical direction.

It can also be understood that the body has no acceleration in the horizontal direction, its displacement can be found as the product of velocity and time.

Comparing the two values of the two cases given in the problem, we can solve it as shown in the diagram below.


In the given problem there are three people at the three corners of a equilateral triangle. The side of the triangle is known to us and each person with a constant velocity approaching the other person. We would like to know after how much time do they meet and where to they meet. The problem is as shown in the diagram below.


Each person is moving towards the other person. We can find that the component of velocity of the second person in the direction of the first person is the cos component and to find effective velocity, we need to add both of them. This is the relative velocity of the approach and it can be expressed as the rate of change of the displacement and it has limits of zero to side of the triangle.

Thus by integrating the equation as shown below, we can find the time after which they are going to meet and the meeting point is the centroid of the triangle.  


It is given in the problem that two train are travelling in the same direction on the same track and they are moving with different speeds. The first driver has noticed it and so he applied breaks and stated decelerating his train. We need to know the value of it so that collision can be avoided.


To avoid the collision, the final relative velocity between the two trains has to be zero. As they are moving in the same direction, relative velocity of approach is the difference between them. We can use this data and apply them in the third equation of motion. To avoid the collision, the distance of separation between has to be greater than the given value of the problem. The solution is as shown in the diagram below.


A problem is given such that time is expressed in terms of displacement as shown in the diagram below. We need to find the acceleration of the body in the given time.


To solve the problem, first we need to find the displacement equation in terms of time by squaring the equation given in the problem. By differentiating the displacement equation, we get acceleration. We need to be careful here as both displacement and velocity are variables with  time. So we need to apply product rule of the differentiation as shown.

We know that the rate of change of velocity is acceleration and rate of change of displacement is velocity. By substituting them in the available equation, we can solve the problem as shown in the diagram below.

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