1D Kinematics, 4 questions Rocket Problem At launch a rocket accelerates upward

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1D Kinematics, 4 questions

Rocket Problem At launch a rocket accelerates upward at a = 2g. After 12 seconds the rocket runs out of fuel and enters freefall. Determine the rocket's maximum height. How long after takeoff does the rocket crash to the ground? Long Ride If you could ride your bike directly from Arcata to the North Pole at a constant speed of v = 4.0 m/s how long would it take in hours? Radar fun Mary is a police officer who has a radar gun that uses metric units. She sees a car stop at a tree (x = 0) and begin to accelerate in the +x direction at time t = 0. After 1 second, she sees they have moved 0.75 m and have velocity 1.5 m/s. After 2 seconds, they are 3 m past the tree and are moving at 3 m/s. After 4 seconds, she sees they have moved 12 m and are moving at 6 m/s. Make a graph of x vs t and a graph of v vs t. Use the same t scale for both graphs (like in lab). Be sure to label your axes and draw things nicely to scale. Use the position data to find the average velocity from t = 0 s to 2 s and from t = 2 s to 4 s. Now use the velocity data to find the average velocity over the same time intervals. What is the acceleration of the car? Is it constant or not? Alien Galileo A scientist on the Planet AA23 is up in a tower, 80 m above the ground. He lets a rock drop from his hand to the ground below. He realizes just as he drops it that he forgot his stopwatch and won't be able to time its fall. After it hits the ground, he inspects the point of impact, and estimates that it was going 22 m/s when it hit. (He's a xenopedologist.) What is the acceleration due to gravity on the surface of Planet AA23? How does this compare with Earth?

Explanation / Answer

1.(a) During first 12 s, acceleration of rocket = a = 2g

Velocity acquired by rocket at end of 12 s = v = ?

v = vo + at where vo = initial velocity = 0 and t = 12s

v = 0 + 12(2g) = 24 g = 24 x 9.81 = 235.44 m/s

To calculate displacement of rocket in first 12 s we can use

y = vot + 0.5at^2

y = 0 + 0.5(2g)(12)^2 = 0.5(2 x 9.81)(12)^2 = 1412.64 m

After 12 s rocket undergoes free fall. It will travel upwards till its final velocity becomes zero. Here its accelerartion is equal to -g

Now displacement of rocket = y'

Final velocity = v' = 0 m/s

Initial velocity = v = 235.44 m/s

Acceleration = -9.81 m/s^2

Using v'^2 = vo^2 + 2ay'

0^2 = 235.44^2+ 2(-9.81)(y')

y' = 2825.28 m

Maximum height reached = y + y' = 1412.64 m + 2825.8 m = 4238.44 m

(b) It displaces by y in 12 s

Then it travels y' in time t'

v' = vo +at'

0 = 235.44 -9.81t'

t' = 24s

Time taken by rocket to fall from its maximum height = t''

where displacement of rocket while falling from maximum height = - maximum height = v''t + 0.5 (-9.81)t''^2

- 4238.44 = 0 - 4.05t''^2

t'' = 32.35 s

Total time taken by rocket to crash after launch = 12 s + t' + t'' = 12s + 24 s + 32.25s = 68.35s

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