4. An aeroplane lands at 216 km/hr and stops after covering a runway of 2 km. Close the
acceleration and the time, in which it comes to rest,
Answers
Answer:
Explanation:
Assuming the acceleration to be uniform, using the equation v^2=u^2+2as (v - final velocity, u - initial velocity, a - acceleration, s-distance) we get
0 = 216^2+(2 x a x 2) ie., a = - 11664 km/hr^2 = - 0.9 m /s^2. The negative sign indicates ‘deceleration’
knowing that acceleration = -11664 km/hr^2, using the equation v = u+at (or t=(v-u)/a) ), we get
t = (0 - 216)/-11664 = 0.0185 hr = 66.66 sec
However, it is to be noted that the above calculation is on the assumption that acceleration is uniform. Usually, in case of decelerations involving aerodynamic drags, the deceleration will not be uniform, ie it will not be a constant during the motion; develerations will be function of the velocity itself, ie, it will depend on the velocity of the aeroplane. The derivation of acceleration as a function of velocity is slightly involved.