Question

QUESTION 2 Earthquakes produce several types of shock waves, two of which are called S and P waves. P waves travel at about 6.5 km/s, and S waves travel at about 3.5 km/s. The delay in arrival times between P waves and S waves provides from the measurement station. If a station records a delay of 33 s, how far from the station was the earthquake (ín kilometers? Use two significant figures in your answer. information about how far away the earthquake epicenter is QUESTION 3 An important piece of landing equipment must be thrown to a ship, which is moving at 45 cm/s towards the shore, before the ship can dock. This equipment is thrown at 15 m/s at 60° above the horizontal from the top of a tower at the edge of the water, 8.75 m above the ship's deck as shown. For this equipment to land at the front of the ship, what distance D from the dock, in meters, should the ship be when the equipment is thrown? Ignore air resistance. QUESTION 4 A subway train starts from rest at a station and accelerates along a straight track at 1.60 m/sf for 14 S. It then runs at constant speed for 70.0 s, and then slows down at a rate of 3.50 m/s2 until it stops at the next station. Find the total distance, in meters, traveled between the stations.

Answer 1

Question 2)

Speed of P waves $v_p=6.5\,km/s$

Speed of S waves $v_s=3.5\,km/s$

Distance of station from earthquake is $d$

Time delay between P and S waves is $t_s-t_p=33\,s$

$v_p=\frac{d}{t_p}\Rightarrow t_p=\frac{d}{v_p}$ , and   $v_s=\frac{d}{t_s}\Rightarrow t_s=\frac{d}{v_s}$

$t_S-t_p=\frac{d}{v_s}-\frac{d}{v_p}$

$33=d\left ( \frac{1}{3.5}-\frac{1}{6.5} \right )$

$33=d\,\frac{3}{3.5*6.5}\Rightarrow d=250.25\,km$

Distance of station from earthquake is $d=250.25\,km$

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Question 3)

The equipment is at height $8.75\,m$,

When the equipment is thrown on to ship, the ship is at a distance of $D$ from the dock.

D= Distance traveled by equipment towards the ship (in horizontal direction) + distance traveled by ship towards the dock.

Time taken by ship to move a distance $x_2$ is $t$. In the same time the equipment moves through a distance $x_1$.

Vertical distance traveled by equipment in time $t$ is  $y_1=-8.75\,m$

Using $y_1=u_yt-\frac{1}{2}gt^2$

$-8.75=15\sin60\degree t-\frac{1}{2}(9.81)t^2$

$-4.905t^2+13\,t+8.75=0$

$t=3.207\,s$

Horizontal distance traveled by equipment in time $t$ is $x_1=u_x t=15\cos60\degree *3.207=24.05\,m$

Distance covered by ship during this time is $x_2=v_s t=0.45*3.207=1.44\,m$

$D=x_1+x_2=24.05+1.44=25.5\,m$

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Question 4)

Train starts with an initial speed  $v_1=0\,m/s$ and accelerates for $t_1=14\,s$ with an acceleration $a_1=1.60\,m/s^2$

Distance covered during this time $S_1=v_1t_1+\frac{1}{2}a_1t_1^2=0+\frac{1}{2}*(1.60)(14)^2=156.8\,m$

$v_2=v_1+a_1t_1=0+1.60*14=22.4\,m/s$

For the next $t_2=70\,s$ the train travels with constant velocity $v_2$

Distance covered during the time $t_2$ is  $S_s=v_2t_2+\frac{1}{2}a_2t_2^2=22.4*70+0=1568\,m$

Then train slows down to $v_3=0\,m/s$with an acceleration of $a_3=-3.50\,m/s^2$ in time $t_3$

Using $v_3^2-v_2^2=2a_3S_3$

Distance covered during this interval is  $S_3=\frac{v_3^2-v_2^2}{2a_3}=\frac{0^2-22.4^2}{2*(-3.50)}=71.68\,m$

Total distance covered between the stations is  is $S=S_1+S_2+S_3=1796.5\,m$

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