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?
Answers
Answer:
20 m/sec
Explanation:
Initial Velocity (u) = 10 m/sec
Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec
Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2
Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2
Initial Velocity (u) = 10 m/secAcceleration (a) = 5 m/sec 2 Distance (s) = 30 m
Initial 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 2
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
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
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 +2as
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
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
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
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
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
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
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
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
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
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
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
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 =
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
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
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
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
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_____________________________
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/sec
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