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kompoz [17]
3 years ago
6

Lindsey started biking to the park traveling 15 mph, after some time the bike got a flat so Lindsey walked the rest of the way,

traveling 4 mph. If the total trip to the park took 5 hours and it was 53 miles away, how long did Lindsey travel at each speed
Physics
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer:

Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

Explanation:

Speed = distance/time

Let the distance that Lindsey biked through be x miles and the time it took her to bike through that distance be t hours

Then, the rest of the distance that she walked is (53 - x) miles

And the time she spent walking that distance = (5 - t) hours

Her biking speed = 15 mph = 15 miles/hour

Speed = distance/time

15 = x/t

x = 15 t (eqn 1)

Her walking speed = 4 mph = 4 miles/hour

4 = (53 - x)/(5 - t)

53 - x = 4 (5 - t)

53 - x = 20 - 4t (eqn 2)

Substitute for X in (eqn 2)

53 - 15t = 20 - 4t

15t - 4t = 53 - 20

11t = 33

t = 3 hours

x = 15t = 15 × 3 = 45 miles.

(53 - x) = 53 - 45 = 8 miles

(5 - t) = 5 - 3 = 2 hours

So, it becomes evident that Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

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

2) Time of flight of the shell: 12.9 s

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Explanation:

1)

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

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y=(u sin \theta)t-\frac{1}{2}gt^2 (1)

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x=(ucos \theta)t

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We can re-write this last equation as

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

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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

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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

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

2)

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

t=\frac{x}{u cos \theta}

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

u=156 ft/s is the initial velocity of the shell

\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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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

#LearnwithBrainly

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