# 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

Expert Answer:

### 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 t^{2}
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

^{2}

### Answered by Thiyagarajan K | 17th Sep, 2021, 12:55: PM

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