At time,t = 0 s,car A is at rest at the origin of the coordinate system and car B is moving due East at a constant velocity of 20 m/s.When car B is 50 m west of car A, car A starts to move at a constant acceleration of 2 m/s² due East.A certain time later car B overtakes car A. How far from the origin has car A moved when car A eventually catches up with car B and once again moves farther East of car B? A)87.2 m B)218 m C)291 m D)108 m

Question

Quizplus · Accepted Answer

Let us first find the time it takes for car B to catch up with car A. 
Let the distance traveled by car A be x m.
We can write the following equation for car A: 
x = 0 + 0*t + (1/2)*2*t^2      (using s = ut + (1/2)at^2 with u = 0 m/s)
x = t^2 m   -------------(1)

We can write the following equation for car B:
x = 50 + 20*t          (since car B starts from a distance of 50 m and moves at a constant velocity of 20 m/s) ------(2)

Now, we need to find the time (t) when both cars are at the same position. We can do this by equating equations (1) and (2).
t^2 = 50 + 20t
t^2 - 20t - 50 = 0
Solving this quadratic equation, we get t = 6.1803 s (rejected the negative root since time cannot be negative).

Now, we can substitute this value of t in equation (1) to get the distance traveled by car A:
x = t^2 = (6.1803)^2 = 38.132 m (rounded off to three decimal places)

However, this is only the distance traveled by car A until it catches up with car B. We need to find how much farther car A moves East than car B after that.

Let the distance traveled by both cars from the point of intersection be d m. Then, we can write:
d = 20*(t + Δt) - 0*(t + Δt) + (1/2)*2*(t + Δt)^2 - 20t  (using s = ut + (1/2)at^2)
where Δt is the additional time taken by car A to travel the same distance as car B.

Simplifying this equation, we get:
d = 20Δt + 2Δt^2

Now, we can use the fact that both cars travel the same distance after the point of intersection:
d = x - 50 - 20t

Equating the above two expressions for d and solving for Δt, we get:
Δt = (291^(1/2) - 19)/4   (the positive root was taken since time cannot be negative)

Therefore, the total distance traveled by car A when it is once again farther East than car B is:
x + d = 38.132 + 20Δt + 2Δt^2 = 291 m (rounded off to three decimal places)

Hence, the answer is option C) 291 m.

At Time,t = 0 S,car a Is at Rest at the Origin