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g100num [7]
3 years ago
13

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

ou plan to stop for dinner. If the fastest you can safely drive is 55 mi/h, what is the longest time you can spend over dinner and still arrive just in time for the meeting? Select one: O a. 0.8 h b. You can't stop at all. c. 2.6 h d. 2.8 h 0
Physics
1 answer:
Lorico [155]3 years ago
5 0

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is  55 mi/h

so if he drives all the to the meeting with max speed then it takes =\frac{450}{55}=8.181 h

and total allowable time is 10.8

Therefore longest time he can spend over dinner is 10.8-8.181 \approx 2.6 hours

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S_a_v_e_r_a_g_e=48km/h

Explanation:

Ok, the average speed can be calculate with the next equation:

S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}   (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

Total\hspace{3}distance=2*d

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

Total\hspace{3}time=t1+t2

From basic physics we know:

t=\frac{d}{S1}

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t2=\frac{d}{S2}

Using the previous information in equation (1)

S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }

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S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}   (2)

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