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Question: A moves with constant velocity u along the x-axis. B always has velocity towards A. After how much t...

A moves with constant velocity u along the x-axis. B always has velocity towards A. After how much time will B meet A if B Moves with constant speed V? What distance will be traveled by A and B?

Explanation

Solution

The question talks about the motion of two bodies or objects A and B in a rectilinear or straight line manner, where the velocity of A started and remains especially in the x axis or direction and B acting opposite to it with a uniform velocity.

Complete step by step answer:
Using the equation of motion which involves initial velocity, final velocity, acceleration and distance travelled or covered.
Appling V=U+atV = U + at
Where V is the uniform velocity of impact from B
U is the uniform velocity of impact from A and a is the acceleration and t is the time taken
For us to calculate t we need to make t subject of the formula or equation by dividing both sides by a, it then become
VU=at VUa=ata t=VUa  V - U = at \\\ \to \dfrac{{V - U}}{a} = \dfrac{{at}}{a} \\\ \to t = \dfrac{{V - U}}{a} \\\
To get the distance travelled by either the objects or bodies in other for them to meet we have to use the second equation of motion, the equation becomes
S=velocity×t+12at2S = velocity \times t + \dfrac{1}{2}a{t^2}, where S is the distance travelled or covered but the acceleration is zero as the velocity is uniform
S=velocity×tS = velocity \times t
The velocity in the above equation will be the average of velocities of A and B
S=U+V2×tS = \dfrac{{U + V}}{2} \times t
Therefore the distance between A and B will be S=(U+V)t2S = \dfrac{{\left( {U + V} \right)t}}{2}

Note: The same kind of motion is seen in cars travelling along a straight path where average speed or velocity, time taken to cover some distances, acceleration along the part of impact as well as other relevant parameters that makes up the motion equation.