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3 years ago

11

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

?

1 answer:

Arisa [49]3 years ago

4 0

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

= 1/2 (6 X 15)

= 1/2 (90)

= 45 meters

Hence, the object traveled 45 meters.

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3 years ago

A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass

stealth61 [152]

Answer:

The shortest distance is $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

The distance between the axial is $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$ behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

For the bicycle not to be thrown over the

Momentum about the back wheel must be zero so

$\sum _B = 0$

=> $mg (26) = ma(40)$

=> $a = \frac{26}{40} g$

Here $g = 32.2 ft/s^2$

So $a = \frac{26}{40} (32.2)$

$a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

$v^2 = u^2 + 2as$

Where $u = 32.26 ft /s$

and $v = 0$ since the bicycle is coming to a stop

$v^2 = (32.26)^2 - 2(20.93) S$

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2 years ago

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

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2 years ago