A jet plane lands with a speed of 85.0 m/s and can accelerate?
with a maximum magnitude of 6.00 m/s2 as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest? s (b) Can this plane land on a small tropical island airport where the runway is 0.800 km long? yes no (c) What is the minimum distance the plane requires to land? km
If a jet plane lands with a speed of 95.0 m/s and can accelerate with a maximum magnitude of 7.00 m/s2.?
A jet plane lands with a speed of 95.0 m/s and can accelerate with a maximum magnitude of 7.00 m/s2 as it comes to rest. What is: a.) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest? b.) Can this plane land on a small tropical island airport where the runway is 0.800 km long? c.) What is the minimum distance the plane requires to land? Lease help me on these few physics problems, thank you.
A jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of -5.80 m/s2 as it comes to re?
A jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of -5.80 m/s2 as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time needed before it can come to rest? s (b) Can this plane land on a small tropical island airport where the runway is 0.800 km long? yes no What is the actual minimum distance needed to stop the plane? km
A jet plane lands with a speed of 100 m/s and can accelerate at a constant rate of -5 m/s2?
a jet plane lands with a speed of 100 m/s and can accelerate at a constant rate of -5 m/s2 as it comes to rest. From the instant the plane touches ground, what is the minimum time needed before coming to a rest? can the plane land with a runway that is .8 km long?
A jet plane lands with a speed of 110 m/s and?
We use equation of motion v^2 - u^2 = 2as v is final velocity = 0 u is initial velocity = 110m/s a = -4.5m/s/s So we calculate distance required by jet to land => 0 - 110*110 = 2 * -4.5 * s => s = 1344m or 1.34 km Since the distance required is more than the 0.9km provided, the plane cannot land in such an airport
Could someone help me with Physics :D ?
For these types of problems there are two equations to memorize: D = D0 + (V0) T + (0.5) A T^2 where T^2 means T squared V = V0 + A T where D = distance traveled at time T D0 = distance at start (T=0) V0 = velocity at start V = velocity at time T A = acceleration of object These equations are for a consistent direction, with positive being in one direction, and negative being in the opposite direction. 1.) V0 = 100 m/s A = - 5.5 m/s2 {negative means the plane is slowing down} We want to make sure V = 0 before D = 0.89 km = 890 m {don't forget to keep consistent units} Using the velocity equation to see how long it would take the plane to stop: V = V0 + A T 0 = (100 m/s) + (- 5.5 m/s2) T T = 18.2 s Using the distance equation to see how far it would take to stop: D = D0 + (V0) T + (0.5) A T^2 D = 0 + (100 m/s)(18.2 s) + (0.5)(-5.5 m/s2)(18.2)^2 D = 909 m = 0.909 km Since the runway is only 0.89 km, the plane will overshoot the runway by 0.909 - 0.89 = 0.02 km 2.) V0 = 0 A = 0.501 m/s2 D = 5.17 m Using the distance equation to see how long it takes the stroller to travel 5.17 m: D = D0 + (V0) T + (0.5) A T^2 5.17 m = 0 + 0 T + (0.5)(0.501 m/s2) T^2 T = 4.54 s Using the velocity equation to find the speed of the stroller at 4.54 s: V = V0 + A T V = 0 + (0.501 m/s2)(4.54 s) = 2.27 m/s 3.) D0 = 0 D = 330 m V0 = 0 {plane starts from rest} V = 120 km/h = (120km/h)(1000 m/km)(1 h/3600s) = 33.3 m/s {keep consistent unit} Using the velocity equation to get an equation relating A and T: V = V0 + A T 33.3 m/s = 0 + A T So A T = 33.3 Using the distance equation to get another equation relating A and T: D = D0 + (V0) T + (0.5) A T^2 330 m = 0 + 0 T + (0.5) A T^2 A T^2 = 660 But since from the velocity equation A T = 33.3, then substituting this into the distance equation: A T^2 = 660 (33.3) T = 660 T = 19.8 s {this is the time it takes to accelerate to 120 km/h} Therefore AT = A (19.8 s) = 33.3 m/s A = 1.68 m/s2 {this is the minimum acceleration the plane needs to take off in 330 m}