Drag Plane

Classical Mechanics Level 2

The above (left) is an F- $117$ Nighthawk being dragging a parachute to reduce the landing run. Also above (right) is the time-velocity graph of this fighter jet landing after flying with a velocity of $v$ . If the velocity before landing doubles, what will be the distance required for the landing? (Assume that the magnitude of air resistance is constant regardless of velocity.)

2 times longer 4 times longer 3 times longer 9 times longer

7 solutions

Abhik Chattopadhyay
Feb 23, 2014

Using v^2=u^2-2 f s (v=0 & u=V) s=V^2/2f using same formula by putting v=0 & u=2V, s'=4V^2/2f So S'=4*S

its simple dude just remember that area under velocity versus time graph displacement or here distance (both are same) since retardation remains same than the slope of the line in the above graph remains same that says that the new triangle made in the new graph will be similar to this one also one of its side is doubled than we can also say that all the sides are doubled hence forth you can understand the concept

Annie Namdev - 7 years, 3 months ago

how can u say that retardation is same. in tne case of drag, retarding force is directly proportionan to squqre of velosity

AVINASH SHROTRIYA - 7 years, 3 months ago

ya its related to velocity but in this problem they have mentioned that air resistance remains same regardless of what the velo is

varun verma - 7 years, 3 months ago

@Varun Verma – yes dear u r correct it is mention in question sorry for inconvinience

AVINASH SHROTRIYA - 7 years, 3 months ago

??? the deceleration keeps constant although its speed double??

Erick Sumargo - 7 years, 3 months ago

lets assume

devansh shringi - 7 years, 3 months ago

Force remains constant (given). So deceleration is also constant

Vinayak k - 7 years, 3 months ago

Hey Eric its up to the one giving the question what he wants to happen and what he doesn't as this is all ideal physics

Annie Namdev - 7 years, 3 months ago

its simply 4 times coz twicing the landing area for double speed doest not match

Ali Javed - 7 years, 3 months ago

Solving via pythagoras is so much simpler...

Oliver Freeman - 7 years, 3 months ago

its double the speed

yug choksi - 7 years, 3 months ago

the v-t graph is starting from the time after the jet lands. hence the parachute is opened at T1 time. when v=2V then also we can assume that parachute opens at T1 time. Since drag force is constant, we would have the same slope. The triangle area then becomes 4 times, but the rectangle (the area before the parachute opens) is only 2 times. hence the total area is more than 2 times and less than 4 times. One cannot find a ratio in this case if the first rectangular area is considered (which I did). And this area is important because it contributes to the distance required for landing.

Sohom Chakrabarty - 7 years, 3 months ago

what does f stand for,i think it should be a i,e acceleration.plz explain

zalmey khan - 7 years, 3 months ago

Santiago Santana
Feb 26, 2014

Distance = $-\frac{V_{0}^{2}}{2a}$

Distance with $v_{0} \times 2$ = $-\frac{(2 \times v_{0})^{2}}{2a} = 4 \times -\frac{v_{0}^{2}}{2a}$

$V_{0}$ : Velocity before landing

$a$ : Aceleration (if air resistance remain constant, same with aceleration)

A Former Brilliant Member
Feb 24, 2014

v^2=u^2-2as ( as 'a' is negative) after landing v=0 therefore u^2=2as hence for 2u distance covered=2^2*s=4s

Jaivir Singh
Apr 18, 2014

IF DEACCELERATION IS CONSTANT THEN STOPPING DISTANCE IS SQUARE OF VELOCITY

Adithyan Ravi
Mar 6, 2014

stopping distance = (v v)/2a first time s1 = (v v)/2a now, v = 2v s2 = (2v 2v)/2a s2 = 4(v v)/2a s2 = 4*s1

DjAbhishek Gowda
Mar 5, 2014

because velocity doubles and length also doubles...............

Engr Rehan Khan Khattar
Mar 2, 2014

simply use the formula of drag force(f) ,drag force is directly proportional to square of velocity (v^2) if velocity become double then v'=2v hence v'^2 =4*v^2 so foce become f '=4f (as all the other factors didn't changes)

Drag Plane

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