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Vadim26 [7]
2 years ago
6

A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

reach the final velocity of 1,000 meters/second.
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
5 0

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

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a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

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p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

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t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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