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Mathematics 21 Online
OpenStudy (anonymous):

A jet plane has a takeoff speed of vto = 75 m/s and can move along the runway at an average acceleration of 1.3 m/s2. If the length of the runway is 2.5 km, will the plane be able to use this runway safely?

OpenStudy (anonymous):

ok so 2 equations \[v=v_o+at\] and \[x=\frac{1}{2}at^2+v_ot+x_o\] the first equation calculates velocity at time t the 2nd equation calculates distance travelled at time t

OpenStudy (anonymous):

ok assuming that initial velocity and initial position are both equal to 0 the take off velocity is 75 and acceleration is 1.3 so determine the time it takes to reach 75 m/s or 75=1.3t solve t

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

it will

OpenStudy (anonymous):

t = 57.69

OpenStudy (anonymous):

once you find t you substitute the values into the position equation and determine how far it will travel so \[x=\frac{1}{2}(1.3)t^2\]

OpenStudy (anonymous):

37.49

OpenStudy (anonymous):

calculate that again i got 2163 m

OpenStudy (anonymous):

Wait thats not it

OpenStudy (anonymous):

You're right. Forgot the squared

OpenStudy (anonymous):

ok now is 2163m less than 2.5km???

OpenStudy (anonymous):

the plane needs to travel 2163 meters in order to reach takeoff velocity and the runway is only 2.5km

OpenStudy (anonymous):

337 meters

OpenStudy (anonymous):

337meters means absolutely nothing unless it asked for how much of the runway wasnt used all you just need to realize is 2.5km is the same as 2500 m and that because the plane only needed 2300m to take off, then there was enough room

OpenStudy (anonymous):

You're absolutely right

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