After landing, a jetliner on a straight runway taxis to a stop at an average velocity of -33.0km/h .
If the plane takes 8.00s to come to rest, what are the plane’s initial velocity and acceleration?
To find the initial velocity and acceleration of the plane, we can use the kinematic equation:
\(v_f = v_i + at\)
Given:
Final velocity (v_f) = 0 km/h (since the plane comes to a rest)
Time (t) = 8.00 s
Average velocity (v_i) = -33.0 km/h
We need to convert the velocities to m/s since the SI unit of velocity is meters per second (m/s). To convert km/h to m/s, we multiply by a conversion factor of 1000/3600.
Converting the average velocity to m/s:
v_i = -33.0 km/h * (1000/3600 m/s per km/h) = -9.17 m/s
Now we can substitute the known values into the kinematic equation:
0 = -9.17 m/s + a * 8.00 s
Simplifying the equation:
-9.17 m/s = -8.00 a
Now we can solve for acceleration (a):
a = (-9.17 m/s) / (-8.00 s)
a = 1.14625 m/s²
Therefore, the plane's initial velocity is -9.17 m/s and its acceleration is 1.14625 m/s².