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At a certain instant of time, a toy car is traveling at a constant speed of v0 to the right, while a distance D ahead is another toy car starting from rest and traveling to the left with constant acceleration a. The cars are moving along the same line, directly toward each other. At what time do they collide, in terms of D, v0 and a?

Respuesta :

Answer:

They collide after [tex]t=\frac{-2u+\sqrt{4u^{2}+8aD}}{2a}[/tex]

Explanation:

The time in which the two cars collide will be equal to the time in which they two cover the distance 'D'.

Let the cars meet after 't' time

Distance covered by toy car travelling at a constant velocity equals

[tex]D_{1}=v_{0}t................(i)[/tex]

Distance that the accelerating car travles in the same time 't' can be obtained from second equation of kinematics as

[tex]D_{2}=ut+\frac{1}{2}at^{2}\\\\D_{2}=\frac{1}{2}at^{2}(\because u=0)[/tex]

Now the condition dictates that

[tex]D_{1}+D_{2}=D\\\\\therefore ut+\frac{1}{2}at^{2}=D[/tex]

Solving the quadratic equation for 't' we get

[tex]at^{2}+2ut-2D=0\\\\\therefore t=\frac{-2u+\sqrt{4u^{2}+8aD}}{2a}[/tex]

We neglect negative value since time cannot be negative.