An Airbus A380 airliner lands at 30m/s Partially loaded, its mass is 480000 kg. The engines app ly reve thrust for 12 s to slow the plane to 25m/s How much thrust did the engines apply?

To find the thrust applied by the engines, we can use the impulse-momentum theorem. The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Mathematically, this can be expressed as:<br /><br />\[<br />F \cdot \Delta t = m \cdot \Delta v<br />\]<br /><br />where:<br />- \( F \) is the force (thrust) applied,<br />- \( \Delta t \) is the time duration over which the force is applied,<br />- \( m \) is the mass of the object,<br />- \( \Delta v \) is the change in velocity.<br /><br />Given:<br />- Initial velocity, \( v_i = 30 \, \text{m/s} \),<br />- Final velocity, \( v_f = 25 \, \text{m/s} \),<br />- Mass, \( m = 480000 \, \text{kg} \),<br />- Time duration, \( \Delta t = 12 \, \text{s} \).<br /><br />First, calculate the change in velocity (\( \Delta v \)):<br /><br />\[<br />\Delta v = v_f - v_i = 25 \, \text{m/s} - 30 \, \text{m/s} = -5 \, \text{m/s}<br />\]<br /><br />Now, substitute the values into the impulse-momentum equation to solve for \( F \):<br /><br />\[<br />F \cdot 12 \, \text{s} = 480000 \, \text{kg} \cdot (-5 \, \text{m/s})<br />\]<br /><br />\[<br />F \cdot 12 = -2400000<br />\]<br /><br />\[<br />F = \frac{-2400000}{12}<br />\]<br /><br />\[<br />F = -200000 \, \text{N}<br />\]<br /><br />The negative sign indicates that the thrust is applied in the opposite direction of the motion, which is expected since it is a decelerating force. Therefore, the magnitude of the thrust applied by the engines is \( 200,000 \, \text{N} \).