vf^2=vi^2+2ad
solve for a
a jetliner, traveling northward,is landing with a speed of 69m/s.once the plane touches down, it has a 750m runway in which speed is reduce to 6.1m/s. compute the avgerage acceleration (magnitude and direction) of the plane during landing.
3 answers
-.042
v^2=u^2 +2as
V=6.1
U=69
A=?
S=750
6.1^2=69^2 + 2 (a) (750)
37.21=4761+1500 a
37.21-4761/1500=a
a=-3.1m/s^2
^(raised to the power of or exponential)
V=6.1
U=69
A=?
S=750
6.1^2=69^2 + 2 (a) (750)
37.21=4761+1500 a
37.21-4761/1500=a
a=-3.1m/s^2
^(raised to the power of or exponential)