Starting from rest an aeroplane takes off after cvering 0.7 km on the runway . If it takes of at 35 ms-1 , calculate (a) the acceleration (b) time for which it moves on the runway .
Answers
Answered by
2
Answer:
Explanation:
Given:- (Initial velocity) v= 0
(Distance) s= 0.7km
= 700m
(Final velocity) v= 35m/s
(a) (Acceleration) a= Using Formula 2as= v²-u²
2a(700)= 35²-0²
1400a= 1225-0
a= 1225÷1400
a= 0.875m/s²
Therefore, the acceleration of the aeroplane is 0.875m/s²
(b)Using Formula v= u+at
35= 0+(0.875)t
t= 35÷0.875
t= 40s
Therefore, the time taken by the aeroplane to move on the runway is 40 seconds.
Hope this helps! =)
Similar questions