answer this PHYSICS MOTION CLASS 9TH
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acceleration =9 upon 5 the distance travelled =250
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Given :
Initial velocity = 36km/h = 36 × 5/18 = 10m/s
Final velocity = 54km/h = 54 × 5/18 = 15m/s
Time = 10 sec
Acceleration = v - u / t
= 15 - 10/10 = 5/10 = 1/2 = 0.5m/s²
Distance = s = ?
From second equation of motion,
s = ut + 1/2at²
= 10 × 10 + 1/2 × 0.5 × 0.5 × 10 × 10
= 100 + 25
= 125m
Hence, Acceleration = 0.5m/s² & Distance = 125m.
Initial velocity = 36km/h = 36 × 5/18 = 10m/s
Final velocity = 54km/h = 54 × 5/18 = 15m/s
Time = 10 sec
Acceleration = v - u / t
= 15 - 10/10 = 5/10 = 1/2 = 0.5m/s²
Distance = s = ?
From second equation of motion,
s = ut + 1/2at²
= 10 × 10 + 1/2 × 0.5 × 0.5 × 10 × 10
= 100 + 25
= 125m
Hence, Acceleration = 0.5m/s² & Distance = 125m.
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