A plane lands on a runway travelling 60m/s north. The plane slows at a rate of 2.0m/s. and comes to a stop after 30 seconds
Calculate the total distance traveledby the plane on the runwayas it was slowing to a stop.
I did
vi=60 m/s
a= -2.0 m/s^s
t=30s
vf=0m/s
d=vit+1/2at^2
d= (60m/s)(30s)+1/2(-2m/s)^2(30s)^2
d=1,800+1,800=3,600
d=3,600
Is this correct?
"The plane slows at a rate of 2.0m/s" , the units should be 2.0 m/s^2
Using basic Calculus:
a = -2
v = -2t + c
when t = 0 (it hits the runway), v = 60
60 = -2(0) + c ----> c = 60
v = -2t + 60
when it stops, v = 0,
0 = -2t+60
t = 30 <--- (we were told that as unnecessary information)
d = -t^2 + 60t + k, but when t=0, d=0 , so k = 0
after 30 sec,
d = -900 + 30*60 = 900 m
thank you for explaining my mistake.
actually, the error was near the end of your solution
in:
d= (60m/s)(30s)+1/2(-2m/s)^2(30s)^2
d=1,800+1,800=3,600
you missed the negative sign in the 2nd term and an arithmetic miscalculation.
should have been:
d= (60m/s)(30s)+1/2(-2m/s)^2(30s)^2
d=1,800 - 900 = 900
so close!
Yes, your approach and calculation are correct. To calculate the total distance traveled by the plane, you can use the equation:
d = vit + 1/2at^2
where:
- d is the distance traveled
- vi is the initial velocity (60 m/s north)
- a is the acceleration/deceleration of the plane (-2.0 m/s^2)
- t is the time taken (30 seconds)
Plugging in the values:
d = (60 m/s)(30 s) + 1/2(-2 m/s^2)(30 s)^2
= 1800 m + 1/2(-2 m/s^2)(900 s^2)
= 1800 m - 900 m
= 900 m
So, the total distance traveled by the plane on the runway as it was slowing to a stop is 900 meters.