a rocket which is moving with velocity 50m/s. eject out its secondary engine and its start to accelerate with 5m/s². for next 10 seconds while entering into space. than it moves with uniform velocity. calculate this uniform velocity and the distance covered in 27 seconds. after ejection of secondary engine.
Asked by ankitparihar70788 | 29th Aug, 2021, 03:38: PM
velocity after 10 s is determined from equation of motion , v = u + ( a t ) ,
where v is velocity after t seconds , u is initial velocity and a is acceleration .
Hence, velocity after 10s , v = 50 + ( 5 × 10 ) = 100 m/s
Hence uniform velocity after acceleration = 100 m/s
Distance S travelled in accelerated motion, S = ( u t ) + (1/2) a t2
S = ( 50 × 10 ) + ( 0.5 × 5 × 10 × 10 ) = 750 m
Distance D travelled in uniform motion for 10s time , D = velocity × time = 100 × 17 = 1700 m
Total distance = ( 1700 + 750 ) m = 2450 m
Answered by Thiyagarajan K | 17th Sep, 2021, 12:55: PM
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