An airplane touches down on the runway with a speed of 70 m/s. It decelerates at a rate of -5 m/s2.
v = vi + at
vi = 70 given
a = -5 given so that determines that
v = 70 - 5t
note:it is not 70 - (-5t)
If you write this down you teacher will not be disappointed, mine wasn’t.
To find the time it takes for the airplane to come to a stop, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s, since the airplane comes to a stop)
u = initial velocity (70 m/s)
a = acceleration (in this case, -5 m/s^2)
t = time (unknown)
Rearranging the equation to solve for time (t), we get:
t = (v - u) / a
Plugging in the given values, we can calculate the time it takes for the airplane to come to a stop:
t = (0 - 70) / -5
t = -70 / -5
t = 14 seconds
Therefore, it takes 14 seconds for the airplane to come to a stop.
v = Vi + a t
Vi = 70 given
a =-5 given
so
v = 70 - 5 t
note: not 70 - (-5t)
a = -5 m/s^2
v = (-5t + k) m/s
when t = 0 , v = 70 m/s
70 = 0 + k ----> k = 70
v = (-5t + 70) m/s
now you can find the velocity at any time after touchdown
but you asked no question.
If the question was, "how long before the plane stops ?" , then
solve
-5t + 70 = 0 for t