Q3. It a Superbike accelerates at 390 km /hr at 1 hr. How much distance will it travelled if the initial velocity is 30km/hr?

Q3. It a Superbike accelerates at 390 km /hr at 1 hr. How much distanc

1.	Q3. It a Superbike accelerates at 390 km /hr at 1 hr. How much distance will it travelled if the initial velocity is 30km/hr?
Answer» 20 m/secExplanation:Initial VELOCITY (U) = 10 m/secInitial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mInitial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third LAW of Motion,Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asInitial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400 Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400 v = 20 m/secInitial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400 v = 20 m/sec_____________________________Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400 v = 20 m/sec_____________________________ANSWER =Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 mFinal velocity (v) = ??Using Newton's Third law of Motion,v 2 =u 2 +2asv 2 =10 2 +2×5×30 v 2 =100+300 v 2 =400 v = 400 v = 20 m/sec_______________________ ANSWER =FINAL VELOCITY = 20 m/secMARK ME AS BRAINLIEST ANSWER

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Q3. It a Superbike accelerates at 390 km /hr at 1 hr. How much distance will it travelled if the initial velocity is 30km/hr?​

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Q3. It a Superbike accelerates at 390 km /hr at 1 hr. How much distance will it travelled if the initial velocity is 30km/hr?