A toy airplane is flying in a straight horizontal (parallel to the ground) path

Question

A toy airplane is flying in a straight horizontal (parallel to the ground) path with a speed of v_i = 23.0 m/s. Suddenly, the airplane's engines stop. Using an x-axis that is horizontal and along the plane's original velocity and a y-axis that is vertical with positive away from the ground, the airplane now has an acceleration given by the following expressions: a_x(t) = t^2/C| a_y(t) = -g[1 - upsilon_x(t)/upsilon_i]| where C = 2831 s^4/m| and upsilon_2(t)| is the x component of the airplane's velocity. The airplane crashes 34.11 seconds after the engines stop. What is the vertical component of the airplane's velocity upsilon_y| at the instant of the crash? When the airplane crashes, what is its impact speed when it strikes the ground? How high up was the airplane before its engines stopped? How far did the airplane travel in the x direction from the time when its engines stopped to the time it crashed?

Explanation / Answer

vx = integration ax*dt

vx = -t^3/3C + vi

vy = integration ay*dt

vy = -g*[ 1 + t^3/(3*C*vi) + 1] *dt

vy = -g*[ t + t^4/(12*C*vi) + t] from t = 0 t = 34.11 s

vy = -9.8*(34.11 + 34.11^4/(12*2831*23) + 34.11 )

vy = -685.53 m/s     <<<=====ANSWER

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vx = -(t^3/3C + vi) = -34.11^3/(3*2831) + 23 = -27.67 m/s

v = sqrt(vx^2+vy^2)

v = sqrt(685.53^2+27.67^2) = 686 m/s     <<<=====ANSWER

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along vertical

vy = integration(ay*dt)

-y = integration (vy*dt) = -g*[ t + t^4/(12*C*vi) + t]*dt from t = 0 t = 34.11 s

-y = -g*(t^2/2 + t^5/(60*C*vi) + t^2/2)

-y = -9.8*(34.11^2 + 34.11^5/(60*2831*23))

h = y = 11518 m    <<<=====ANSWER

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x = integration(vx*dt) = -t^4/12C - vi = -34.11^4/(12*2831) -23 = -62.85 m

distance travelled = 62.85 m <<<=====ANSWER

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