# Thread: Derivaties in relation to velocity, speed,etc

1. ## Derivaties in relation to velocity, speed,etc

Suppose that the distance an aircraft travels along a runway befoer takeoff is given by $\displaystyle D=\frac {10t^2}{9}$ , where D is measured in meters from the starting point and t is measured in seconds from that time the brakes are released If the airplane will become airborne when its speed reaches 200 km/h, how long will it take to become airborne, and what distance will it have traveled by that time?

Phew.

Okay, I think

1st find velocity (derivative of distance)
$\displaystyle v(t)= \frac {20t}{9}$

Then set $\displaystyle 200= \frac {20t}{9}$

Then solve for t. T= 90 h

2nd plug 90 for t in the original equation.

and solve for t.

t= 9000 km/h

Yes? NO? Maybe so?

Thank you!

2. v(t) = dD/dt = 20t/9

200*1000/3600 = 20t/9
t = 25 seconds

it takes 25 second before airborne

the distance = 10(625)/9= 694,444444 meters

3. Originally Posted by Singular
v(t) = dD/dt = 20t/9

200*1000/3600 = 20t/9
t = 25 seconds

it takes 25 second before airborne

the distance = 10(625)/9= 694,444444 meters
Where did 1000/3600 come from? How did you get 1000/3600?

The rest I understand.

4. Originally Posted by Truthbetold
Where did 1000/3600 come from? How did you get 1000/3600?

The rest I understand.
Hello,

the speed was given in $\displaystyle \frac{km}{h}$.

1 km = 1000 m
1 h = 3600 s

Thus $\displaystyle 200 \frac{km}{h} = 200\cdot \frac{1000m}{3600 s}$