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`        Plxxxx see image and ans que fast,,,.....its urgent......`
5 months ago

1717 Points
```							we can easily solve such probs using relative motion.consider the frame of reference as the car1 having accn= 15m/s^2. the other we call car2then, a21= 5 – ( – 15)= 20. note here sign of 15 is negative coz accns are in the opposite drections. and the vector formula a21=a2 – a1 is even given in ncert (the most fundamental formula).similarly, we find the velocity of car2 wrt car1v21= 10 – ( – 20)= 30m/s.now s=400, u=30, a=20so s= ut+1/2at^2 you can find t=5skindly approve:)
```
5 months ago
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### Course Features

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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions