Car A𝐴 starts from a point O𝑂 and moves with constant velocity 9m/s9𝑚/𝑠. After 2s2𝑠, another car B𝐵 begins its journey from O𝑂 and follows car A𝐴. If car B𝐵 starts from rest and moves under constant acceleration 4m/s24𝑚/𝑠2, after how much time and at what distance from O𝑂 the cars meet?
Question
Car A𝐴 starts from a point O𝑂 and moves with constant velocity 9m/s9𝑚/𝑠. After 2s2𝑠, another car B𝐵 begins its journey from O𝑂 and follows car A𝐴. If car B𝐵 starts from rest and moves under constant acceleration 4m/s24𝑚/𝑠2, after how much time and at what distance from O𝑂 the cars meet?
Solution
Let's solve this step by step:
Step 1: Find the time when car A and car B meet
We know that car A has a constant velocity of 9 m/s and car B has a constant acceleration of 4 m/s². We also know that car B starts 2 seconds after car A.
The distance covered by car A and car B will be the same when they meet.
The distance covered by car A is given by the formula: d = vt where v is the velocity and t is the time.
The distance covered by car B is given by the formula: d = ut + 0.5at² where u is the initial velocity, a is the acceleration and t is the time.
Since car B starts 2 seconds after car A, the time for car B will be (t-2).
Setting the two equations equal to each other gives us: 9t = 0.54(t-2)²
Solving this equation gives us t = 4.5 seconds.
Step 2: Find the distance from O when the cars meet
We can find the distance by substituting t = 4.5 seconds into the equation for the distance covered by car A: d = 9*4.5 = 40.5 meters
So, the cars will meet 4.5 seconds after car A starts, at a distance of 40.5 meters from point O.
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