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Grade: 11
        
Plxxxx see image and ans que fast,,,.....its urgent......
one month ago

Answers : (1)

Aditya Gupta
1347 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 car2
then, a21= 5 – ( – 15)= 20. note here sign of 15 is negative coz accns are in the opposite drections. and the vector formula a21=a– ais even given in ncert (the most fundamental formula).
similarly, we find the velocity of car2 wrt car1
v21= 10 – ( – 20)= 30m/s.
now s=400, u=30, a=20
so s= ut+1/2at^2 you can find t=5s
kindly approve:)
one month ago
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