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Vikki [24]
3 years ago
11

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

Physics
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

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Answer:

a) 4.9 s

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Explanation:

Hello!

To solve this question we need to make use of the equations of motion of both the motorcycle xm(t) and the car xc(t) at t=5

Let us consider the position of the motorcycle at t=5 as the origin, that is:

xm(t+5) = vt + (1/2)at^2

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We are looking for the time t' when the position of the car and the motorcycle are the same:

xm(t'+5)=xc(t'+5)

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Since we are considering the origin of the cooordinate system at the position when the motorcycle starts to accelerate, the distance travelled by the motorcycle until it catches the car is given by:

xm(t'+5)= vt' + (1/2)at'^2

xm(9.89898s) = (22 * 9.89898 + 2.5 * 9.89898^2)m

xm(9.89898s)= 167.777... m

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