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marusya05 [52]
3 years ago
8

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

e the parachute is going down.) It comes to rest at 0 meters/second after 30 seconds. What's the average acceleration of the shuttle during this time span?
Physics
1 answer:
dexar [7]3 years ago
6 0

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

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Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

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The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

                             S = ut + ½ gt²

Let tₓ be the time where the two stones  meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

                            x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

                S - x = U tₓ - ½ gtₓ²  (∵ the motion of the stone is in opposite direction)

Adding these two equations,

                      x + (S - x) = U tₓ

                               S = U tₓ

                               200 = 50 tₓ

∴                                  tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s   

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