To solve the problem, let's break it down into three parts:
1. Determining the distance covered by cars A and B as a function of time:
For uniformly accelerated rectilinear motion, we can use the following equation to calculate the distance covered:
distance = initial velocity * time + (1/2) * acceleration * time^2
For car A:
The initial velocity is 0 m/s, and the acceleration is 4 m/s².
So, the distance covered by car A at time t is:
d¹(t) = 0.5 * 4 * t^2 = 2t^2
For car B:
The initial velocity is 0 m/s, and the acceleration is -2 m/s² (opposite direction to car A).
So, the distance covered by car B at time t is:
d²(t) = 0.5 * -2 * t^2 = -t^2
2. Deducing the time of meeting for the two cars:
To find the time of meeting, we need to set the distances covered by both cars equal to each other:
2t^2 = -t^2
Simplifying the equation:
2t^2 + t^2 = 0
3t^2 = 0
Since the equation equals zero, the only solution is t = 0. This means that the two cars meet at the starting point at t = 0.
3. Considering car B in uniform rectilinear motion with a speed of 10 m/s:
If car B is moving at a constant speed of 10 m/s, it means its acceleration is 0 m/s². Therefore, the equation for car B's distance covered becomes:
d²(t) = initial velocity * time = 10 * t = 10t
Now, we can answer the preceding questions using this new equation for car B.
1. The distance covered by car A at time t is still d¹(t) = 2t^2.
The distance covered by car B at time t is now d²(t) = 10t.
2. The time of meeting for the two cars can be found by setting the distances equal to each other:
2t^2 = 10t
Simplifying the equation:
2t^2 - 10t = 0
2t(t - 5) = 0
From this equation, we have two solutions:
t = 0 (the initial meeting point)
t = 5 seconds (when the two cars meet again after 5 seconds)
Therefore, the cars meet again after 5 seconds.
Please note that the distances calculated above are in terms of t, the time elapsed since t = 0.
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