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3 years ago

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A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

y of 5.47 m/s. She then continues running at this constant velocity. How long does the runner take to travel 165 m?

1 answer:

Stells [14]3 years ago

6 0

Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)= $1.31 m/s^2$

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

$5.47=0+1.31\times t$

$t=\frac{5.47}{1.31}=4.17 s$

Distance traveled during this time is

$s=ut+\frac{1}{2}at^2$

$s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m$

So remaining distance left to travel with constant velocity=153.57 m

thus time $=\frac{distance}{speed}$

$t_2=\frac{153.57}{5.47}=28.07 s$

Total time=28.07+4.17=32.24 s

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what happens to the current in a circuit if the resitance of the components in the circuit is increased

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1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

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where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

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At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

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Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

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In order to find how long the mortar shell remain in the air, we can use the equation:

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x = 1322 ft is the final position of the shell when it strikes the hill

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$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

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Instead, the vertical component of the velocity is given by

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$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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