Starting from rest at a height of 0 m and time of 0 seconds, the rocket is ignit

Question

Starting from rest at a height of 0 m and time of 0 seconds, the rocket is ignited and the acceleration begin. The first acceleration stage lasts for 1.83 s and accelerates the fireworks to un upward velocity of 6.0 m/s. The second acceleration stage lasts for 3.66 seconds and accelerates the fireworks to an upward velocity of 10.80 m/s. Determine the acceleration of the fireworks during the second fule stage. Determine the distance the rocket travels upward (height above gound) during the first stage of motion. Determin the distance the rocket travels upward during the second stage of motion. When the rocket blows up, what time does the rocket change directions and start to head back downward toward the ground?

Explanation / Answer

For stage 2 :

Vi = initial velocity = 6 m/s

Vf = final velocity = 10.80 m/s

t = time = 3.66 sec

a2 = acceleration

Using the equation

a2 = (Vf - Vi) / t = (10.80 - 6) / (3.66) = 1.31 m/s2

For first stage :

Vi = initial velocity = 0 m/s

Vf = final velocity = 6 m/s

t = time = 1.83 sec

a1 = acceleration

Using the equation

a1 = (Vf - Vi) / t = (6 - 0) / (1.83) = 3.28 m/s2

Distance travelled in first stage :

Y1 = Vi t + (0.5) a1 t2 = (0) (1.83) + (0.5) (3.28) (1.83)2 = 5.49 m

Distance travelled in second stage :

Y2 = Vi t + (0.5) a2 t2 = (6) (3.66) + (0.5) (1.31) (3.66)2 = 30.7 m

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