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tester [92]
3 years ago
6

A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0

m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Physics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

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A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
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Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

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