An aeroplane accelerates down at a runway at 3.20m/s^2 for 32.8s until is finally lifts off the ground. Determine the distance traveled before take off . WHO KNOWS THE ANSWER AND SOLUTIONS THEN ONLY YOU WILL GIVE ANSWER ANOTHER WISE I WAS REPORT HIS OR HER ANSWER . THANK YOU.
Answers
Answered by
1
Given, the initial velocity = 0 m /s.
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
Anonymous:
Thank you.
plzzz mark me as brainlist
but the answer is given d=1720m
Answered by
0
I also solved this question previously,
Here in this questions Equation of motion will help
Now it is given thatu=0m/sa=3.20 m/s²t=32.8sand we have to find S
By using equation of motionS=ut + 1/2 at²S=(0x32.8) + (1/2)x3.2x 32.8²S=3442.68 m
So, the airoplane covered the distance of 3442.68 m before the final lift off
#Prashant24IITBHU
Here in this questions Equation of motion will help
Now it is given thatu=0m/sa=3.20 m/s²t=32.8sand we have to find S
By using equation of motionS=ut + 1/2 at²S=(0x32.8) + (1/2)x3.2x 32.8²S=3442.68 m
So, the airoplane covered the distance of 3442.68 m before the final lift off
#Prashant24IITBHU
Similar questions